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Number System- 2: Remainders
by Total Gadha - Saturday, 22 November 2008, 02:01 PM
 

Suppose the numbers N 1 , N 2 , N 3 .. give quotients Q 1 , Q 2 , Q 3 .. and remainders R 1 , R 2 , R 3 ..., respectively, when divided by a common divisor D.

Therefore N 1 = D × Q 1 + R 1 ,

N 2 = D × Q 2 + R 2 ,

N 3 = D × Q 3 + R 3 .. and so on.

Let P be the product of N 1 , N 2 , N 3 ...

Therefore, P = N 1 N 2 N 3 .. = (D × Q 1 + R 1 )(D × Q 2 + R 2 )(D × Q 3 + R 3 )..

= D × K + R 1 R 2 R 3 ... where K is some number ---- (1)

In the above equation, only the product R 1 R 2 R 3 ... is free of D, therefore the remainder when P is divided by D is the remainder when the product R 1 R 2 R 3 ... is divided by D.

Let S be the sum of N 1 , N 2 , N 3 ...

Therefore, S = (N 1 ) + (N 2 ) + (N 3 ) +...

= (D × Q 1 + R 1 ) + (D × Q 2 + R 2 ) + (D × Q 3 + R 3 )..

= D Ã— K + R 1 + R 2 + R 3 ... where K is some number--- (2)

Hence the remainder when S is divided by D is the remainder when R 1 + R 2 + R 3 is divided by D.

Examples:

  1. What is the remainder when the product 1998 × 1999 × 2000 is divided by 7?

Answer: the remainders when 1998, 1999, and 2000 are divided by 7 are 3, 4, and 5 respectively. Hence the final remainder is the remainder when the product 3 × 4 × 5 = 60 is divided by 7.

Answer = 4

  1. What is the remainder when 2 2004 is divided by 7?

2 2004 is again a product (2 × 2 × 2... (2004 times)). Since 2 is a number less than 7 we try to convert the product into product of numbers higher than 7. Notice that 8 = 2 × 2 × 2. Therefore we convert the product in the following manner-

2 2004 = 8 668 = 8 × 8 × 8... (668 times).

The remainder when 8 is divided by 7 is 1.

Hence the remainder when 8 668 is divided by 7 is the remainder obtained when the product 1 × 1 × 1... is divided by 7

Answer = 1

  1. What is the remainder when 2 2006 is divided by 7?

This problem is like the previous one, except that 2006 is not an exact multiple of 3 so we cannot convert it completely into the form 8 x . We will write it in following manner-

2 2006 = 8 668 × 4.

Now, 8 668 gives the remainder 1 when divided by 7 as we have seen in the previous problem. And 4 gives a remainder of 4 only when divided by 7. Hence the remainder when 2 2006 is divided by 7 is the remainder when the product 1 × 4 is divided by 7.

Answer = 4

  1. What is the remainder when 25 25 is divided by 9?

Again 25 25 = (18 + 7) 25 = (18 + 7)(18 + 7)...25 times = 18K + 7 25

Hence remainder when 25 25 is divided by 9 is the remainder when 7 25 is divided by 9.

Now 7 25 = 7 × 7 × 7 3 .. (8 times) × 7 = 343 × 343 × 343... (8 times) × 7.

The remainder when 343 is divided by 9 is 1 and the remainder when 7 is divided by 9 is 7.

Hence the remainder when 7 25 is divided by 9 is the remainder we obtain when the product 1 × 1 × 1... (8 times) × 7 is divided by 9. The remainder is 7 in this case. Hence the remainder when 25 25 is divided by 9 is 7.

Some Special Cases:

2.1A When both the dividend and the divisor have a factor in common.

Let N be a number and Q and R be the quotient and the remainder when N is divided by the divisor D.

Hence, N = Q × D + R.

Let N = k × A and D = k × B where k is the HCF of N and D and k > 1.

Hence kA = Q × kB + R.

Let Q 1 and R 1 be the quotient and the remainder when A is divided by B.

Hence A = B × Q 1 + R 1 .

Putting the value of A in the previous equation and comparing we get-

k(B × Q 1 + R 1 ) = Q × kB + R --> R = kR 1 .

Hence to find the remainder when both the dividend and the divisor have a factor in common,

    • Take out the common factor (i.e. divide the numbers by the common factor)
    • Divide the resulting dividend (A) by resulting divisor (B) and find the remainder (R 1 ).
    • The real remainder R is this remainder R1 multiplied by the common factor (k).

Examples

  1. What the remainder when 2 96 is divided by 96?

The common factor between 2 96 and 96 is 32 = 2 5 .

Removing 32 from the dividend and the divisor we get the numbers 2 91 and 3 respectively.

The remainder when 2 91 is divided by 3 is 2.

Hence the real remainder will be 2 multiplied by common factor 32.

Answer = 64

2.1B THE CONCEPT OF NEGATIVE REMAINDER

15 = 16 × 0 + 15 or

15 = 16 × 1 - 1.

The remainder when 15 is divided by 16 is 15 the first case and -1 in the second case.

Hence, the remainder when 15 is divided by 16 is 15 or -1.

--> When a number N < D gives a remainder R (= N) when divided by D, it gives a negative remainder of R - D.

For example, when a number gives a remainder of -2 with 23, it means that the number gives a remainder of 23 - 2 = 21 with 23.

EXAMPLE

  1. Find the remainder when 7 52 is divided by 2402.

Answer: 7 52 = (7 4 ) 13 = (2401) 13 = (2402 - 1) 13 = 2402K + (-1) 13 = 2402K - 1.

Hence, the remainder when 7 52 is divided by 2402 is equal to -1 or 2402 - 1 = 2401.

Answer: 2401.

2 .1C When dividend is of the form a n + b n or a n - b n :

rule

EXAMPLES

  1. What is the remainder when 3 444 + 4 333 is divided by 5?

Answer:

The dividend is in the form a x + b y . We need to change it into the form a n + b n .

3 444 + 4 333 = (3 4 ) 111 + (4 3 ) 111 .

Now (3 4 ) 111 + (4 3 ) 111 will be divisible by 3 4 + 4 3 = 81 + 64 = 145.

Since the number is divisible by 145 it will certainly be divisible by 5.

Hence, the remainder is 0.

  1. What is the remainder when (5555) 2222 + (2222) 5555 is divided by 7?

Answer:

The remainders when 5555 and 2222 are divided by 7 are 4 and 3 respectively.

Hence, the problem reduces to finding the remainder when (4) 2222 + (3) 5555 is divided by 7.

Now (4) 2222 + (3) 5555 = (4 2 ) 1111 + (3 5 ) 1111 = (16) 1111 + (243) 1111 .

Now (16) 1111 + (243) 1111 is divisible by 16 + 243 or it is divisible by 259, which is a multiple of 7.

Hence the remainder when (5555) 2222 + (2222) 5555 is divided by 7 is zero.

  1. 20 2004 + 16 2004 - 3 2004 - 1 is divisible by:

(a) 317 (b) 323 (c) 253 (d) 91

Solution: 20 2004 + 16 2004 - 3 2004 - 1 = (20 2004 - 3 2004 ) + (16 2004 - 1 2004 ).

Now 20 2004 - 3 2004 is divisible by 17 (Theorem 3) and 16 2004 - 1 2004 is divisible by 17 (Theorem 2).

Hence the complete expression is divisible by 17.

20 2004 + 16 2004 - 3 2004 - 1 = (20 2004 - 1 2004 ) + (16 2004 - 3 2004 ).

Now 20 2004 - 1 2004 is divisible by 19 (Theorem 3) and 16 2004 - 3 2004 is divisible by 19 (Theorem 2).

Hence the complete expression is also divisible by 19.

Hence the complete expression is divisible by 17 × 19 = 323.

2.1D When f(x) = a + bx + cx 2 + dx 3 +... is divided by x - a

rule

EXAMPLES

  1. What is the remainder when x 3 + 2x 2 + 5x + 3 is divided by x + 1?

Answer: The remainder when the expression is divided by (x - (-1)) will be f(-1).

Remainder = (-1) 3 + 2(-1) 2 + 5(-1) + 3 = -1

  1. If 2x 3 -3x 2 + 4x + c is divisible by x - 1, find the value of c.

Since the expression is divisible by x - 1, the remainder f(1) should be equal to zero.

Or 2 - 3 + 4 + c = 0, or c = -3.

2 .1E Fermat's Theorem

rule

EXAMPLE

  1. What is the remainder when n 7 - n is divided by 42?

Answer: Since 7 is prime, n 7 - n is divisible by 7.

n 7 - n = n(n 6 - 1) = n (n + 1)(n - 1)(n 4 + n 2 + 1)

Now (n - 1)(n)(n + 1) is divisible by 3! = 6

Hence n 7 - n is divisible by 6 x 7 = 42.

Hence the remainder is 0.

2.1F Wilson's Theorem

rule

EXAMPLE

  1. Find the remainder when 16! Is divided by 17.

16! = (16! + 1) -1 = (16! + 1) + 16 - 17

Every term except 16 is divisible by 17 in the above expression. Hence the remainder = the remainder obtained when 16 is divided by 17 = 16

Answer = 16

2.1G TO FIND THE NUMBER OF NUMBERS, THAT ARE LESS THAN OR EQUAL TO A CERTAIN NATURAL NUMBER N, AND THAT ARE DIVISIBLE BY A CERTAIN INTEGER

To find the number of numbers, less than or equal to n, and that are divisible by a certain integer p, we divide n by p. The quotient of the division gives us the number of numbers divisible by p and less than or equal to n.

EXAMPLE

14. How many numbers less than 400 are divisible by 12?

Answer: Dividing 400 by 12, we get the quotient as 33. Hence the number of numbers that are below 400 and divisible by 12 is 33.

15. How many numbers between 1 and 400, both included, are not divisible either by 3 or 5?

Answer: We first find the numbers that are divisible by 3 or 5. Dividing 400 by 3 and 5, we get the quotients as 133 and 80 respectively. Among these numbers divisible by 3 and 5, there are also numbers which are divisible both by 3 and 5 i.e. divisible by 3 x 5 = 15. We have counted these numbers twice. Dividing 400 by 15, we get the quotient as 26.

Hence the number divisible by 3 or 5 = 133 + 80 - 26 = 187

Hence, the numbers not divisible by 3 or 5 are = 400 - 187 = 213.

16. How many numbers between 1 and 1200, both included, are not divisible by any of the numbers 2, 3 and 5?

Answer: as in the previous example, we first find the number of numbers divisible by 2, 3, or 5. from set theory we have

n(AUBUC) = n(A) + n(B) + n(C) - n(A intersectn B) - n(B intersectn C) - n(A intersectn C) + n(A intersectn B intersectn C)

n(2U3U5) = n(2) + n(3) + n(5) - n(6) - n(15) - n(10) + n(30)

--> n(2U3U5) = 600 + 400 + 240 - 200 - 80 - 120 + 40 = 880

Hence number of numbers not divisible by any of the numbers 2, 3, and 5

= 1200 - 880 = 320.

I shall have to end here and leave the rest of it for my CBT Club students. I shall cover some problems based on this in the CBT Club this week.

 

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Re: Number System- 2: Remainders
by piyush jain - Tuesday, 6 January 2009, 10:26 AM
 

hi,

this is a good one article.

piyush

Re: Number System- 2: Remainders
by Mayur Bhambhani - Wednesday, 14 January 2009, 04:28 AM
 

Hi TG Sir

Respect!!

Sir the article is named as number system-2

Was there a number system 1 and if it was where is it?

I don't wanna miss ne of ur articles

plzz let me know

 

Re: Number System- 2: Remainders
by Nikhil Rajagopalan - Thursday, 29 January 2009, 10:33 AM
 

Hi TG,

Thank you for this is excellent article for starters.

Regards,

Nikhil

 

 

Re: Number System- 2: Remainders
by Arpita Bhattacharya - Sunday, 1 February 2009, 05:04 PM
  if R=[(0.15)55+(0.25)55]/[(0.15)54+(0.25)54]
1.R>0.20
2.R>0.1
3.R=4
4.R<0.25
I am new here plz reply......
Number System- 2: Remainders
by prasanna samal - Tuesday, 3 February 2009, 02:46 PM
  Can anyone help in finding the solution of a problem.
The problem is find the remainder when 444^444^444/7.
Re: Number System- 2: Remainders
by shivam mehra - Monday, 9 February 2009, 12:56 AM
  good question  444^444^444/7.

euler theorem
7,444 coprime to each other

phi(7) = 6
therefore 444^6 % 7 = 1..........(a)

power 444^444 has to be broken in 6k+r form

444%6 =0   444^444=6k+0=6k

question reduces to 444^6k........(b)

from (a) and (b)
remainder is 1^k i.e 1 only


Re: Number System- 2: Remainders
by abhinav singh - Monday, 9 February 2009, 02:38 PM
  one more method i think works well for such questions.

its 444^444^444

for the time being take the no to be 444^k

its remainder with 7 will be:- (444/7)^k ie (3)^k or(-4)^k

now k = 444^444

therefore the remainder is (-4)^444^444

again taking one 444 and other as x we get

R = ((-4)^444)^x)
now 4^4 = 64 and leaves 1 as remiander
therefore it becomes R = 1^111^x therefore remainder is 1
With practise one can reduce the time taken to solve this type of sums.
Re: Number System- 2: Remainders
by shivam mehra - Monday, 9 February 2009, 10:52 PM
  plz explain ::after this

again taking one 444 and other as x we get

R = ((-4)^444)^x)
now 4^4 = 64 and leaves 1 as remiander
therefore it becomes R = 1^111^x therefore remainder is 1
With practise one can reduce the time taken to solve this type of sums.
Re: Number System- 2: Remainders
by abhinav singh - Wednesday, 11 February 2009, 02:51 PM
  sorry i made a mistake...if we take 4^3 = 64 the remainder with 7 will be one....now it is nothing but (1) ^ 148 ^ 444 ie = 1. i think now u wil get it...my mistake sorry..
Re: Number System- 2: Remainders
by shivam mehra - Friday, 13 February 2009, 12:02 PM
  complete approach

444^444^444 mod 7

444 mod 7 = 3 or -4

(-4)^even no. ==4 ^even no.

even no. = 444^444
444= 6k+0
(k=74)
original number reduces to equivalent to

4^6k  mod 7 =1


Re: Number System- 2: Remainders
by vikas sharma - Monday, 16 February 2009, 05:28 PM
 

hi can anybody brief it .i cudnt get it

 

 

Re: Number System- 2: Remainders
by abhishek rai - Saturday, 18 April 2009, 12:45 PM
  Nice one......better than any cat prep book i have consulted.....
doing grt job sir...
Re: Number System- 2: Remainders
by pratish raj - Saturday, 18 April 2009, 07:01 PM
  check for R=[(0.15)^1+(0.25)^1]/[(0.15)^0+(0.25)^0]
this is equal to 0.2
now when u increase both the numerator and denominator, the fraction will definitely increase.
hence, R>0.20
Re: Number System- 2: Remainders
by Hardik Shah - Friday, 1 May 2009, 08:13 PM
  [3^101]/77 find the reminder?
Re: Number System- 2: Remainders
by Hardik Shah - Friday, 1 May 2009, 08:49 PM
  [17^36+19^36]/111 what is the reminder?
Re: Number System- 2: Remainders
by Hardik Shah - Saturday, 2 May 2009, 11:33 PM
  Got the ans of 3^101 /77 .
ans is 47.
Re: Number System- 2: Remainders
by venkat s - Monday, 4 May 2009, 10:57 PM
 

Please help me solving this problem please provide the soultion as well

Q) 1 and 8 are 2 natural numbers for which 1+2+3+....+n  is a perfect square

which number is 2nd third and 4th such number .

 

Re: Number System- 2: Remainders
by Varun Agrawal - Thursday, 28 May 2009, 06:28 PM
  @venkat 49 and 288.
n.(n+1)/2 = x^2
=> n/2 and (n +1) both are perfect squares. So look for an odd square and half of its next number should also be a square 
Re: Number System- 2: Remainders
by nishant saxena - Saturday, 30 May 2009, 07:59 PM
 

hi sir

i wann aknow in ques 16 why u hav added 80 instead of subtracting it?????

 

 

Re: Number System- 2: Remainders
by abhishek bindal - Sunday, 31 May 2009, 11:21 PM
  @ hardik.. is the ans for your ques is 52.. do let me knoe
Re: Number System- 2: Remainders
by Deepika Khandelwal - Tuesday, 2 June 2009, 10:05 AM
  hi varun,
how to check :--
AB36 (A & B is any no. ) is perfect square this type of questions
please provide solution

Re: Number System- 2: Remainders
by Deepika Khandelwal - Tuesday, 2 June 2009, 12:42 PM
  hardik..how u got answer..... Got the ans of 3^101 /77 .
ans is 47.
could you please explain
Re: Number System- 2: Remainders
by Priyesh Tungare - Tuesday, 2 June 2009, 05:25 PM
  Hi,
In the question:
  1. What the remainder when 2 96 is divided by 96?
y the common factor between 296 and 96 was 32? Can anybody please tell me y this two number were taken?

Also in the question where we have to find the remainder when 22004 is divided by 7, y did we took 8668? Please help me in this matter. sad

Thanks,
Priyesh Tungare
Re: Number System- 2: Remainders
by Deepika Khandelwal - Wednesday, 3 June 2009, 11:58 AM
  if we      see that 96 is also even so definetly there will be common something
so 296     =25 291  and 96 is divisible by 32 so we have taken 32
as a common factor
I think I am able to solve your doubt


Re: Number System- 2: Remainders
by Deepika Khandelwal - Wednesday, 3 June 2009, 12:08 PM
  And for 22004 in this case our effort is to make reminder 1
and if we divide 8 by 7 it gives reminder 1 it will be easy to solve the problem
(23)668   equals to 8668
Re: Number System- 2: Remainders
by VIKAS RAI - Thursday, 4 June 2009, 12:38 AM
  Hi
Cn you plz tel hw did u gt 47. ans shud be 26. please explain.
Re: Number System- 2: Remainders
by VIKAS RAI - Thursday, 4 June 2009, 06:51 AM
  sorry guys i'v got 47.i was doing it in a diff. way
Re: Number System- 2: Remainders
by amit bhoir - Friday, 5 June 2009, 11:09 AM
 

 

Hi Deepika this is how i got .slightly lengthy process.

3^101 /77can be written as

(3^100 * 3)/77   = (81^25 *3) /77 =((77 +4)^25 * 3)/77

it can be now written as

(4^25 * 3)/77 = (4^24 *4 *3)/77

                     =  ((77-13)^8 *4 *3)/77

                 = ((-13)^8 * 4*3)/77

                 =(13^8 *4 *3)/77

                  =((154 +15)^4 *4 *3)/77

                  =(15^4 *4 *3)/77

                   =((231 - 6)^2 *4 *3)/77

                    =(6^2 * 4 *3)/77

           thus remainder comes 47

I hope I have not  done error.Anybody with less lengthy process please reply.

Re: Number System- 2: Remainders
by dindo saha - Friday, 5 June 2009, 11:02 PM
  better method
(444^444)/7=(3^444)/7=(9)^222/7=(2)^222/7=(8)^74/7=1
best of luck
(soln courtesy to my roomate at IIT kgp hostel abhisek)
Re: Number System- 2: Remainders
by dindo saha - Friday, 5 June 2009, 11:15 PM
  (200!)/(100!)^2 remainder?
200!=1*2*3*4...........*200/(1*2*3*4...100)*(1*2*.....100)=
101*102*..200/(100!)
now 101*102*......200 has 100 elements .since product of consecutive n terms divisible by n! thus remainder=0
Re: Number System- 2: Remainders
by William Wallace - Sunday, 7 June 2009, 03:09 PM
  Hi Deepika,

Your approach is right. 296=25x3

so prob reduces to finding remainder of 291/3.Now applying Euler's theorem we get remainder as 2.

Now we multiply 2 with 25(common factor) and hence the final ans is 64.
Re: Number System- 2: Remainders
by Deepika Khandelwal - Monday, 8 June 2009, 10:07 AM
  what is the remainder of
2147   divided by 11 ???
 can anybody give solution how to solve this.
Re: Number System- 2: Remainders
by Deepika Khandelwal - Monday, 8 June 2009, 10:21 AM
  If N is divided by D gives a reminder 4 then 9N is divided by same divisor
gives 7 reminder find the value of N ???
 
how to solve this type of questions.....

Can anyone give solution for this also ...

Re: Number System- 2: Remainders
by TG Team - Monday, 8 June 2009, 08:32 PM
  210 = 1 mod11
=> 2147 = 214x10 + 7 = 27 mod11 = 128 mod11 =  7 mod11smile
Re: Number System- 2: Remainders
by TG Team - Monday, 8 June 2009, 08:38 PM
   Given that
 N = aD + 4
9N = bD + 7
=> D > 7
from eqn 1, we get 9N = 9aD + 36
comparing with eqn 2
36 = cD + 7
=> cD = 36 - 7 = 29
=> D = 29 as it cannot be 1. smile
Re: Number System- 2: Remainders
by Deepika Khandelwal - Tuesday, 9 June 2009, 10:49 AM
  it is not clear kamal....
how it came...?
36 = cD + 7
Number System- 2: Remainders
by srinivasan ravi - Tuesday, 9 June 2009, 06:45 PM
  hi evryone...pls help me solve this problem.

find the remainder when 1!+2!+3!+........+49! is divided by 7.


thanks smile
Re: Number System- 2: Remainders
by saloni ghosh - Tuesday, 9 June 2009, 08:52 PM
 

Let S = 1!+2!+3!+4!+...+49! = 2!+3!+4!+5!+(6!+1)+7!+...+49!

since terms from 7!+..are multiples of 7

also according to wilson's theorem 6!+1 is divisible by 7

therefore S= 2!+3!+4!+5! + 7*(some positive integer) = 152 + 7* (integer)

remainder = 5

 

Re: Number System- 2: Remainders
by Ashwin A - Wednesday, 10 June 2009, 06:49 AM
  find the remainder for ( 5^99) / 66
Re: Number System- 2: Remainders
by srinivasan ravi - Wednesday, 10 June 2009, 09:42 AM
  thanks saloni..but what if the divisor is 17 instead of 7.??is there any shortcut??
Re: Number System- 2: Remainders
by TG Team - Wednesday, 10 June 2009, 12:20 PM
 

Ø(66) = 66(1 - 1/2)(1 - 1/33) = 32.

=> 532 = 1 mod66.

=> 599 = 532x3 + 3 = (532)3 x 53 = 53 mod66 = 125 mod66 = 59 mod66. smile

Re: Number System- 2: Remainders
by prerak valdiya - Wednesday, 10 June 2009, 12:23 PM
 

to your question 2^147 /11

i think ....

2^147 = 2^2 * 2 ^145 = (2^5)^71*2^2

(32)^71 *4 /11 = (33-1)^71 * 4/11 = (-1)^71 * 4 /11 = 40

but 40 /11 =  7  is the final remainder... B-<>

 

Re: Number System- 2: Remainders
by saloni ghosh - Wednesday, 10 June 2009, 02:42 PM
  @ ravi : I don't know of any shortcut. If you find any shortcut , Please do let me know.
Re: Number System- 2: Remainders
by Mushabber Zaidi - Wednesday, 10 June 2009, 09:08 PM
  can ny1 tell me where r the new posts by TG..m new to this site..
Re: Number System- 2: Remainders
by yashaswini kotresh - Wednesday, 10 June 2009, 10:32 PM
  are u sure ur answer is correct kamal..Can u elaborate ur technique.didnt get it.
Re: Number System- 2: Remainders
by Veena Binya - Wednesday, 10 June 2009, 11:39 PM
  Reckon Ø(66) has to be 66(1 - 1/2)(1 - 1/3)(1 - 1/11) ?
Re: Number System- 2: Remainders
by shivam mehra - Thursday, 11 June 2009, 01:03 PM
  [17^36+19^36]/111 what is the reminder?
Re: Number System- 2: Remainders
by saloni ghosh - Thursday, 11 June 2009, 07:29 PM
 

111=3*37

17^36=1(mod 37)

19^36=1(mod 37)

17^2=1(mod 3)  therefore , 17^36=1 (mod 3)

||ly 19^36=1(mod 3)

so, (17^36+19^36)=2(mod 37)

    (17^36+19^36)=2(mod 3)

hence (17^36+19^36)=2(mod 111)

Remainder = 2

 

Re: Number System- 2: Remainders
by Sid manihar - Saturday, 4 July 2009, 05:11 PM
  What is the remainder of the 2000^1000 when divided  by 13
Re: Number System- 2: Remainders
by Jyoti Nath - Sunday, 5 July 2009, 11:39 AM
  3 is the answer
Re: Number System- 2: Remainders
by debashis saha - Monday, 6 July 2009, 10:18 AM
  hi everyone
i am stuck with a question can anyone help me out with the solution ...
Q> A man writes the natural number starting from 789...........till the 192th digit
  find the last digit? please explain your answer in details
Re: Number System- 2: Remainders
by debashis saha - Monday, 6 July 2009, 10:21 AM
  Q> A person has written natural numbers side by side and it has 10,000digits .
Divide the number by 8 to get the remainder .What will be the remainder ?
find the last three digit also ?
Re: Number System- 2: Remainders
by debashis saha - Monday, 6 July 2009, 10:24 AM
  8927 digits starts with 6 any two digit number is divisible by 17 or 23.what will be the 5293rd digit ?
Re: Number System- 2: Remainders
by Saurav Modi - Monday, 6 July 2009, 09:46 PM
 

Hi Debashis !!

see if it works...

7,8,9 are 3 digits..

if we will write all two digited numbers from 10 to 99 it will be 100 numbers .i.e, 200 digits. which will make it 203 digited number ending with .....90919293949596979899.

we need only 192 digits so remove 11 digits from the end,which will give the last digit as 9.

Re: Number System- 2: Remainders
by debashis saha - Monday, 6 July 2009, 11:23 PM
  thank you sourav it was a gr8 help
Re: Number System- 2: Remainders
by Namit Midha - Tuesday, 7 July 2009, 12:15 AM
 

can you pls help me with this question

N=123456789101112...............9989991000

 

Find remainder when number formed by the first thousand digits (from left) of Nis divided by 16

 

A)7          B)9          C) 11         D)13      E)15

 

 

Re: Number System- 2: Remainders
by Software Engineer - Tuesday, 7 July 2009, 10:16 AM
  Namit,

16 is a factor of 10^4, therefore divide the last four digits of N to get the remainder.
Answer = 1000 mod 16 = 8.

Refer Divisibility and Bases for more details.

- SE
Re: Number System- 2: Remainders
by Atishay Jian - Tuesday, 7 July 2009, 12:32 PM
  hi namit....
the answer is 13
the last four digits of d no formed by the first 1000 digits is 3693 which divided by 16 gives a remainder of 13.....
Re: Number System- 2: Remainders
by debashis saha - Tuesday, 7 July 2009, 01:27 PM
  sourav

 can you tell me how 10-99 you are getting 100numbers in my calculations it is 93 numbers
Re: Number System- 2: Remainders
by debashis saha - Tuesday, 7 July 2009, 01:33 PM
  hello TG sir

 can you please tell me answers of the above questions they are toubling me a lot .....
Re: Number System- 2: Remainders
by debashis saha - Tuesday, 7 July 2009, 01:34 PM
  q> 232 to 756 how many 5 ,s are there ? 
Re: Number System- 2: Remainders
by Software Engineer - Tuesday, 7 July 2009, 01:37 PM
  Oh! "number formed by the first thousand digits (from left) of N" and I divided the last four digits of N.


Re: Number System- 2: Remainders
by crab anandita - Wednesday, 8 July 2009, 08:30 PM
 

111=3*37

17^36=1(mod 37)

19^36=1(mod 37)

17^2=1(mod 3)  therefore , 17^36=1 (mod 3)

||ly 19^36=1(mod 3)

so, (17^36+19^36)=2(mod 37)

    (17^36+19^36)=2(mod 3)

hence (17^36+19^36)=2(mod 111)

Remainder = 2

but since remainders are 2 in both cases so ans is 2

what if remainder of mod 37 is something else say 5]

n 2 only for mod 3 then??

what would be the final answer??

Re: Number System- 2: Remainders
by crab anandita - Thursday, 9 July 2009, 01:24 PM
 

please solve this ques..

What is the remainder when 683 + 883 is divided by 49 ??

Re: Number System- 2: Remainders
by shivam mehra - Friday, 10 July 2009, 10:29 PM
 
Q)
3 101 mod 77
77(1-1/7)(1-1/11)=60

phi (77)=60
3 60 mod 77 = 1
remaining is 3 41 mod 77

(3 4)10 X 3=4 10 X3  =16  X 3 X 256X 256 mod 77 =47


Q)
2 147 mod 11

2 10 mod 11= 1
2 140 mod 11 =1
2 7 =128 mod 11= 7

Q)
find the remainder when 1!+2!+3!+........+49! is divided by 7.

1+2+6+24+120+720 =1+2+6+3+1+6
==5

Q)What is the remainder of the 2000^1000 when divided  by 13

=50 1000=11 1000 =121 500 = 4 500 =16 250 =3 250 = (3 3 )83 3=3

Q> A man writes the natural number starting from 789...........till the 192th digit
  find the last digit? please explain your answer in details

192 digits means 192/3=64 nos. therefore 789+64-1=852
-1 is done coz 789 is included


Q)
A person has written natural numbers side by side and it has 10,000digits .
Divide the number by 8 to get the remainder .What will be the remainder 
find the last three digit also ?

1to 9              = 9X1=9        
10 to 99          = 90X2=180
100 to 999      = 900X3=2700
10000-2889=7111
7111%4=1777+3/4
1....9........99......999.........2776 277
1    9       189    2889                 10000
last 3 digits 277
rem by 8=5


Q)

What is the remainder when 683 + 883 is divided by 49 ??

i think its 0

(7-1) 83 +(7+1) 83
last term 7 2 -1+7 2 +1===0
Re: Number System- 2: Remainders
by shivam mehra - Friday, 10 July 2009, 10:46 PM
  now solve

1   7^7^7 mod 13

2   25^25^25 mod  9

3   37 ^47^ 57 mod 16
Re: Number System- 2: Remainders
by Rohit Arora - Saturday, 11 July 2009, 11:55 AM
  1 7^7^7 mod 13

phi (13) = 12

WE need to find the remainder of 7^7 by 12...
(-5)^ 7
= (-5)^ 6 . (-5)
(25)^3 . (-5)
1.(-5)

Hence 7^7 = 12k -5 or 12k + 7

7^7 / 13...
(-6)^ 7
(36)^3 . (-6)
(-3)^3. (-6)
-27. -6
+ 27. 6
1.6

Ans= 6

Ok a lil too lengthy any suggestions



2. 25^25^25 mod 9
This one is easy
phi(9) = 6

25^25= (24+1)^ 25 = 6k + 1

=> 25^ (6k +1)
= 25^6k . 25
= 25/9
25 = 3.9 - 2
r= -2 or + 7


3. 37 ^47^ 57 mod 16

= 5 ^ 47^ 57

phi (16)= 8

47= 8k - 1
47^ 57 / 8 ... r = -1

5^ 7 / 16
= 125 X 125 X 5
= -3 X -3 X 5
= 45
= -3 or + 13... Ans


Correct me if i'm wrong.. thanx
Re: Number System- 2: Remainders
by Rohit Arora - Saturday, 11 July 2009, 12:07 PM
  What is the remainder when 683 + 883 is divided by 49 ??

i think its 0

(7-1) 83 +(7+1) 83
last term 7 2 -1+7 2 +1===0

Hey shivam.. i dunno if im wite but the asnwer wont b 0

A^83 + B^83
A= 7-1
B= 7+1

In both the cases when u expand it using the binomial theorem,
all the terms are divisible by 49 except the last 2 terms of both A and B.
Last two terms are

A = 83C82.7.(-1)^ 82 + 83C83(-1)^83
83.7 - 1


B= 83C82.7.(1)^ 82 + 83C83(1)^83
83.7 + 1

Hence R = (2 X 83 X 7)/49
= (2 X 83)/7
= 2 X 6
= 12
12 X 7= 84
84-49 =
35... Ans

Please tell me if there are any mistakes
Re: Number System- 2: Remainders
by Romil Vaghela - Saturday, 11 July 2009, 01:23 PM
 

Hi all !

A good one !

Proved quite helpful to me !

Re: Number System- 2: Remainders
by Priyank Agrawal - Monday, 13 July 2009, 11:33 AM
 

plz solve

Remainder when 1^39 + 2^39 + 3^ 39 +...+ 12^39 is divided by 39.

Re: Number System- 2: Remainders
by piyush jain - Tuesday, 14 July 2009, 05:23 PM
 

plz tell me how to solve follwing,

21^875/17 what is the reminder?

piyush

Re: Number System- 2: Remainders
by Rajarshi Guha - Wednesday, 15 July 2009, 04:12 PM
  Hii Priyank.The answer would be zero.

My approach is 1^39+2^39+3^39+....+12^39 can be written as

(1+2+3+...+12)(1^38+2^38+....12^38)
the first bracket is nothing but sum of the first 12 nos.
12(12-1)/2=78
Since 78 is completely divisible by 39,therefore the above expression would leave a rem=0 when divided by 39..
Re: Number System- 2: Remainders
by Rajarshi Guha - Wednesday, 15 July 2009, 04:16 PM
  Hi Piyush..

21^875/17
First find the rem of 21 /17.It is 4
therefore its nothing but 4^875/17
Now we know that 4^2=16
therefore,(16^437)*4/17
ie (-1)*4/17
ie 16*4/17
ie 64/17 therefore rem is 13
Re: Number System- 2: Remainders
by Rajarshi Guha - Wednesday, 15 July 2009, 04:24 PM
  Hi rohit,

Dont you have the answers given for this?
My approach is

683+883/49 can be written as
156/49
or 54*29/49
or 5*29/49
or 145/49 which gives -2 as the rem
therefore rem is 49-2=47.

Plz post the answer
Re: Number System- 2: Remainders
by aman sukhija - Wednesday, 15 July 2009, 09:52 PM
 

hi Rohit!

given below is appreciable effort from your side.

could you please explain how u figure out phi (13) = 12, phi(9) = 6, phi (16)= 8. in short what is this phi and how to calculate it.

 

1 7^7^7 mod 13

phi (13) = 12

WE need to find the remainder of 7^7 by 12...
(-5)^ 7
= (-5)^ 6 . (-5)
(25)^3 . (-5)
1.(-5)

Hence 7^7 = 12k -5 or 12k + 7

7^7 / 13...
(-6)^ 7
(36)^3 . (-6)
(-3)^3. (-6)
-27. -6
+ 27. 6
1.6

Ans= 6

Ok a lil too lengthy any suggestions



2. 25^25^25 mod 9
This one is easy
phi(9) = 6

25^25= (24+1)^ 25 = 6k + 1

=> 25^ (6k +1)
= 25^6k . 25
= 25/9
25 = 3.9 - 2
r= -2 or + 7


3. 37 ^47^ 57 mod 16

= 5 ^ 47^ 57

phi (16)= 8

47= 8k - 1
47^ 57 / 8 ... r = -1

5^ 7 / 16
= 125 X 125 X 5
= -3 X -3 X 5
= 45
= -3 or + 13... Ans

Re: Number System- 2: Remainders
by harsh maru - Wednesday, 12 August 2009, 06:42 AM
 

Hi,

6^83 + 8^83 when divided by 49, the remainder will not be zero.

consider: (7-1)^83 + (7+1)^83

everyterm, except the last two terms will be divisible by 49, therefore consider only the last two terms.

=83(7.1^82)-1+83(7.1^82)+1

=14.83/49 = 166/7 = 5

Regards,

Harsh

Re: Number System- 2: Remainders
by animesh chandan - Friday, 14 August 2009, 06:37 PM
 

hi prasanna,

hope u got ur answer but there is one more solution of ur question apart from using eulers theoram(infact it is the most suited sol)

   444^444^444/7

we can write it as 444^444*444*444....444times

divide it by 7

we get 3^444*444.....444times

we can write it as(3^3)^148*444*444.... 443times

tat is 27^148*444*444*444...443/7times 

by negetive remainders we get (-1)^148*444*444

since 148*444*444... is even so we get

1^148*444*444...../7   tat is 1/7,

so the remainder is 1

hope so ull find it easy. 

Re: Number System- 2: Remainders
by animesh chandan - Friday, 14 August 2009, 10:32 PM
  hardik,(late again)
the answer of [3^101]/77 is 47
Re: Number System- 2: Remainders
by abhishek tripathi - Saturday, 15 August 2009, 02:14 AM
  well u dnt know what u have given 2 da junta this will help dem fetch a hell lots of marks in various exams i have a liitle doubt when 25 raised to 25 divided by seven actulaay i cant understnad why u multplied 7 to the power 3 eight times..and den multiplied by 7 ....and one more thing n,(n+1),(n-1)how come it is divisble by factorial 3 plz expalin
Re: Number System- 2: Remainders
by saurav verma - Thursday, 20 August 2009, 02:41 AM
 

hi abhishek,

actually it 25^25/9 not divided by 7

so wen we divide 25/9 we get 7 as a remainder

so we can write it as 7^25

and 7^3+ 7^3+..............(8 times=8*3=24)+7

so it will equal to 7^25

 

and about ur 2 nd dought

take any value for n

eg. n=2

den n,(n+1),(n-1)=2,1,3

so 2*1*3=6 divided by 6 equal to 3!

and it is always divided by 3

i think ur dought is clear now.

 

Re: Number System- 2: Remainders
by vikas sharma - Thursday, 3 September 2009, 03:41 PM
 

Hi All

 

pls give me sol of following ques.Ans is 7

How many positive integers less than 100 can be written as the sum of 9 consecutive positive integers‌

Re: Number System- 2: Remainders
by vikas sharma - Friday, 4 September 2009, 02:05 PM
 

Hi Shivam /all

pls tell me following ques , its answer is 320.

 Let f be a factor of 120, then the number of positive integral solutions of xyz = f is

Re: Number System- 2: Remainders
by abhishek tripathi - Friday, 4 September 2009, 06:59 PM
  HI SIR I WAS STUCK AS I WAS SOLVING ONE QUEST OF ARUN SHARMA DAT GOES LIKE DIS WAY (1FACTORIAL RAISED TO THE POWER 3)+(2 FACTORIAL RAISED TO THE P0WER 3)+(3 FACTORIAL RAISED TO THE POWER 3)+(4 FACTORAIL RAISED TO THE POWER3)+..........................+(1152 FACTORIAL RAISED TO THE POWER 3)IS DIVIDED BY 1152 FACTORIAL ....i dnt know how 2 type it but anyone can solve it and show da solution plz rep asap
Re: Number System- 2: Remainders
by Pallav Jain - Tuesday, 15 September 2009, 11:38 AM
 

can anyone plz tell me the answer wid Solution..?

6^83+8^83/49

(6 Raise to powe 83) + (8 Raise to powr 83) Divided By 49 rEMAINDER?

Re: Number System- 2: Remainders
by Pallav Jain - Tuesday, 15 September 2009, 12:17 PM
 

Hi Vikas,

This question has little bit Mistake QUESTION IS.

What is the Sum of all the numbers Less then 100, that can be written AS the Sum of 9 Consecutive Integers. ( HERE SUM OF 9 CON. INT shud be 100 ) tongueout

ANSWER IS :- 7

Sol:- 1-9, 2-10, 3-11, 4-12, 5-13, 6-14, 7-15 these 7 terms will give the Sum 100 ( 9 +con Integers. )

Re: Number System- 2: Remainders
by ROHIT K - Tuesday, 15 September 2009, 01:35 PM
  Hi pallav,

6^83+8^83/49

phi(49)=42 
==> (6^42+8^42)%49=1


Now,

(6^83+8^83)%49=(6^-1 +8^-1)%49= -14

-14+49===>35(Remainder) smile

Re: Number System- 2: Remainders
by Pallav Jain - Tuesday, 15 September 2009, 08:16 PM
 

Hi Rohit,

Well thnx for reply but Sorry I cud'nt understand ur Method. Can u plz Explain it lil More or Any other Method.

If you will scroll down this page u will get to know tht many students hv asked this question nd hv posted different answers.

Plz explain it ......

Thanks

Re: Number System- 2: Remainders
by ROHIT K - Wednesday, 16 September 2009, 11:08 AM
  Sorry..I was lazy and ate steps tongueout .

I will try to explain using two methods:
1. Binomial Theorem
2. Euler Theorem

1. Binomial Theorem Method (I think everyone knows this one)

(6^83+8^83)%49= ((7-1)^83 + (7+1)^83)%49

Expand (a-b)^n and (a+b)^n

You will notice that all the terms would be divisible by 49 except last 2,
since all the terms would have power of 7 more than 2 except the last 2 which would be 7^1 and 7^0 (not divisible by 49).

So we r left with 83C82 * 7 - 1 + 83C82 * 7 + 1(last two terms of both expansion)

The question now boils down to (2*83*7) % 49

Divide by 7 ==> Question becomes (2*83)%7 viz. 166%7=5

Since we divided by 7, we have to multiply the intermediate remainder by 7.

Therefore,

The Final Remainder = 5 * 7 = 35




2. Euler Method

Calculating Euler Totient of 49 written as phi(49).

phi(49)=49*(1-(1/7))=6*7=42

Therefore we have,

(6^42k)%49=1 and (8^42k)%49=1 (By Euler Theorem) k is any natural number


Now,

Question is (6^83+8^83)%49

Adjusting powers to suit the above method.

((6^84 * 6^(-1)) + (8^84 * 8^(-1))) %49

6^84%49 =1 And 8^84 %49=1 (By Euler Theorem)

Hence the question becomes:

((6^-1) + (8^-1))%49

=(8*48^-1 + 6*48^-1)mod49 (by taking L.C.M and solving)

=48^-1(8+6)mod49
=48^-1*14mod49
=-1^-1*14mod49 ..................(48%49=-1)
=14*-1mod49
=-14mod49
=35............(49-14=35)

Hence the Final Remainder=35

Hope this explanation is clear.smile
Re: Number System- 2: Remainders
by Suresh S - Monday, 21 September 2009, 01:51 PM
  HI sir, nice article
Re: Number System- 2: Remainders
by abhishek tripathi - Tuesday, 22 September 2009, 03:00 AM
  hiiiiii thx sourav ya my doubht is clear now can u help me in solving dis one .. 
find the remainder..
(1!)3+(2!)3+(3!)3+.............+(1152!)3/(1152!)
cheers n thx sourav once again...
tg rockz...:p
Re: Number System- 2: Remainders
by varun sood - Tuesday, 22 September 2009, 01:02 PM
  very awesome reply brthr
Re: Number System- 2: Remainders
by ROHIT K - Tuesday, 22 September 2009, 08:33 PM
  Hi abhi,

(1!)3+(2!)3+(3!)3+.............+(1152!)3/(1152!)

is the remainder 0?

will post approach if answer is confirmed..smile



Rohit
Re: Number System- 2: Remainders
by abhishek tripathi - Wednesday, 23 September 2009, 02:42 AM
  hiii rohit gud try but dude i want 2 say dat me sorry its 1152 in reamiander instead of 1152! n answer is 225 .
cheers
Re: Number System- 2: Remainders
by ROHIT K - Wednesday, 23 September 2009, 10:53 AM
  Hi abhi...

I think, the remainder you are talking about as 1152 is actually the divisor..

If it is so, thanks for simplifying the problem.

I will continue the problem taking 1152 as the divisor.

Q. (1!)3+(2!)3+(3!)3+.............+(1152!)3/(1152)

Ans:

Divisor = 1152 = 27 * 32

(1!)3 = 1 % 1152 = 1

(2!)3 = 8 % 1152 = 8

(3!)= 216 % 1152 = 216

(4!)3 = (24)3 = (2* 33 ) % 1152 = 0 cool

By observation, for higher factorial value you will find that remainder with 1152 is 0.................(since power of 2 and 3 will be more than 7 and 2 respectively)

OR

You can say, 4! will occur @ higher factorial value which is already divisible by 1152 hence the remainder with 1152 will be 0.

Hence the remainder = 1 + 8 + 216 = 225 evil

Hope this helps..smile

Rohit


Re: Number System- 2: Remainders
by Tarun Raheja - Thursday, 24 September 2009, 01:14 AM
 

probably better soln:-

2^147/11 = (2^5)^29*4/11 = (-1)29*4/11 = -4/11

Final Remainder = 11-4 = 7

Re: Number System- 2: Remainders
by Tarun Raheja - Thursday, 24 September 2009, 01:36 AM
 

hi atishay

can u pls tell me hw u found the last 4 digits of that number?

 

Re: Number System- 2: Remainders
by Pankaj Kumar - Sunday, 27 September 2009, 08:06 PM
  Hi, Please tell method to solve- what is the remainder when 9091  is divided by 15 (as 15 and 90 are not co prime)

Thanks in Advance

Re: Number System- 2: Remainders
by Pankaj Kumar - Sunday, 27 September 2009, 09:14 PM
 

Hi Debashis !!

I think its like this..

7,8,9 are 3 digits..

if we will write all two digited numbers from 10 to 99 it will be 90(99-9) numbers .i.e,180 digits. which will make it 183 digited number , remaining 9 digit (192-183) .....ending with 100101102

 will give the last digit as 2.

Re: Number System- 2: Remainders
by jyoti verma - Monday, 28 September 2009, 11:47 PM
  I think its remainder is 0
Re: Number System- 2: Remainders
by Vipul Gupta - Tuesday, 29 September 2009, 02:56 AM
  Hi Pankaj ans is 0.
Re: Number System- 2: Remainders
by Kunal Parekh - Wednesday, 7 October 2009, 12:19 AM
 

Dear TG Sir,

Thank you very much.

 

Re: Number System- 2: Remainders
by Romil Vaghela - Thursday, 5 November 2009, 02:21 PM
 

Hi,

Can any1 explain this ????

Re: Number System- 2: Remainders
by amit amit - Thursday, 5 November 2009, 10:38 PM
  What is the Sum of all the numbers Less then 100, that can be written AS the Sum of 9 Consecutive Integers. ( HERE SUM OF 9 CON. INT shud be 100

0= -4+ -3+..   +0+  ..+3+4
9=  -3+ -2+....+5
18= -2+-1+....+6
27=-1+ 0+......+7
36= 0+1+.......+8
45= 1+2+........+9
....
....
99= 7+8+........+15

so, answer is =
0+9+18+27+36+45+54+63+72+81+90+99
= 9(1+2+3+4+5+6+7+8+9+10+11)
=9* (11*12/2)
=9* 11*6
=594   (here all no.s .which r less than 100  r considered  whole no.s)
Re: Number System- 2: Remainders by sumit verma
by sumit verma - Thursday, 5 November 2009, 11:29 PM
  hiii deepika..
2^147=[(2^5)^29]*4
        =(32^29)*4
now when it will be divided by 11 i.e;(32/11) will give a negative remainder of 1 i.e;(-1)
now when (-1) will be raised to a power of 29 ,it will give (-1)
hence the final expression will be [(-4)/11]
which will result in a final remainder of 7
Re: Number System- 2: Remainders
by Sagar Thakur - Wednesday, 18 November 2009, 01:54 AM
  Hello Sir can u explain me Fermat  theorem more broadly.
Re: Number System- 2: Remainders
by Pallav Jain - Wednesday, 18 November 2009, 04:17 AM
  Problem {--}
A = {179, 180, 181,…..,360}. B is a subset of A such that sum of no two elements of B is divisible by 9. The number of elements in B cannot exceed????

a)102    81)   c)101   d)82   e)  None of these

Cheers !!!!!!!!!!

Pallav Jain
Re: Number System- 2: Remainders
by amar goswami - Wednesday, 18 November 2009, 01:59 PM
  number  sum   (additional sum req)      Numbers      Total
1+7+9 = 17       +1/10                        (181,.. 352)  = 20
1+8+7 = 16       +2/11                        (182,.. 353)  = 20
1+8+6 = 15       +3/12                        (183,.. 354   = 20
1+8+5 = 14       +4/13                        (184,.. 355   = 20
1+8+4 = 13       +5/14 =  NA
1+8+3 = 12       +6/15 =  NA
1+8+2 = 11       +7/16 =  NA
1+8+1 = 10       +8/17 =  NA
1+8+0 = 9         +9/18 =                     (189,.. 360)  = 20

above derivation is based upon the fact that if number is divisible by 9 then sum of the digits of the number should be divisible by 9; for example sum of 179 is 17 hence it requires a number having sum of its digits as 1 or 10 in order to be divisible by 9.
Total such numbers =20*5 =100

360-179 = (181 +1)-100 = 82 
Re: Number System- 2: Remainders
by Richa Chauhan - Monday, 4 October 2010, 12:51 PM
  Brilliantly complied and beautifully explained !! Thanks TG
Re: Number System- 2: Remainders
by arun reddy - Saturday, 30 April 2011, 12:28 PM
  please can u explain this solution
Re: Number System- 2: Remainders
by nancy sharma - Saturday, 11 June 2011, 04:42 PM
  can someone help me in solving this

find the 28383rd term of series: 1234567891011121314....
Ans: 3


Re: Number System- 2: Remainders
by TG Team - Sunday, 12 June 2011, 09:35 AM
 
Hi Nancysmile

First observe that there are 9 1-digit numbers which take 9*1 = 9 digits, 90 2-digit numbers which take 90*2 = 180digits, 900 3-digit numbers which take 900*3 = 2700 digits, 9000 4-digit numbers which take 9000*4 = 36000 digits.
So it can be concluded easily that our required digit will be some digit in a 4-digit number. Now we need to find out that which digit of what 4-digit number, right?
That can be easily done if I make all numbers 4-digit ones by inserting some 0's before them. Like insert 3 0's before every 1-digit number (i.e. 3*9 = 27 0's), 2 0's before every 2-digit number (i.e. 2*90 = 180 0's) and 1 zero before every 3-digit number (i.e. 1*900 = 900 0's) to meke every number a 4-digit one.

Now we have inserted 27 + 180 + 900 = 1107 zeros in all so we need to find out 28383 + 1107 = 29490th digit in the following sequence: 0001 0002 0003 0004 0005 0006 0007 0008 0009 0010 0011 0012 .....

Which can be easily found as 2nd digit of 7373 i.e. 3 (because 29490 = 4*7372 + 2).

Please read all the above points carefully. I am sure that you'll be able to get the logic. smile

Kamal Lohia
Re: Number System- 2: Remainders
by shubham singhal - Tuesday, 21 June 2011, 12:14 AM
  plz solve this prob:

what is the remainder when(999/99)^9?
plz state the method and the logic too..finding too much diff in solving this...
Re: Number System- 2: Remainders
by TG Team - Tuesday, 21 June 2011, 02:06 PM
  Hi Shubham smile

Please re-write the question clearly that which number is being divided by whom as there can be many (mis)interpretations of your statement.

Kamal Lohia
Re: Number System- 2: Remainders
by shubham singhal - Wednesday, 22 June 2011, 10:37 PM
  the question is:

what is the remainder when 999^9 is divided by 99^9?

Both den and num. are raised to the power 9.
hvng diff in finding the logic needed to solve the above prob since the dnominator is also raised to a very high power.
looking forward to the solution along with the method.
Re: Number System- 2: Remainders
by vishal jain - Tuesday, 28 June 2011, 08:35 PM
  find the remainder.

3^5^7^9 divided by 41.
Re: Number System- 2: Remainders
by vishal jain - Tuesday, 28 June 2011, 08:37 PM
  also find the remainder of 5^25^125^3125 divided by 11.
Re: Number System- 2: Remainders
by TG Team - Wednesday, 29 June 2011, 01:26 PM
 
Hi Vishal smile

It seems that you have not tried the problems yourself. If you would have given them a try, you could have easily cracked them.

See 34 = 81 = 41(2) - 1. So 38 will leave remainder 1 when divided by 41.
That means now we need to find the remainder of power of 3 i.e. 57^9 with 8.
Now we know that 5² = 25 = 8(3) + 1 i.e. 57^9 is of the form 52k + 1 and leave remainder 51 = 5 when divided by 8 and is of the form 8k + 5.

So finally we need to find the remainder when 38k + 5 is divided by 41 or the remainder when 35 is divided by 41 which is -3 or 38. smile

Hope you can try the next one on your own.
Kamal Lohia 
Re: Number System- 2: Remainders
by vishal jain - Wednesday, 29 June 2011, 04:42 PM
  Thanks for yr help....as a matter of fact I tried this problem by using eruler equation but couldn't reached final answer......anyways keep doing great work...!!!
Re: Number System- 2: Remainders
by TG Team - Wednesday, 29 June 2011, 04:58 PM
 
Vishal smile

In your second question, the numbers 5 and 11 are coprime. So we can apply Euler's Theorem which says 5phi(11) = 510 = 1 mod11.

But remember this phi(11) = 10 may not be the smallest power of 5 which when divided by 11 gives 1 as remainder. I mean that there may be a smaller power of 5 which when divided by 11 gives remainder 1. Here important point to ponder over is that the smallest power, if it exists other than 10 (i.e. phi(11)), will be a divisor of 10 (i.e. phi(10)).

On manual checking, we can easily deduce that 55 = 3125 = 1 mod11.

In given number we have the number of the form 55k only which will also give final remainder as 1 only. smile

Kamal Lohia 
Re: Number System- 2: Remainders
by Debashish ... - Monday, 4 July 2011, 05:02 PM
  [17^36+19^36]/111 what is the reminder?

Is the answer to the above problem is 2 ?
Re: Number System- 2: Remainders
by Rajasekaran Rajaram - Monday, 4 July 2011, 06:32 PM
  Debashish,

Yes the remainder will be 2.

111 = 37 * 3

Φ(111)  = LCM[Φ(37) ,Φ(3)]= 36

1736 mod 111 = 1
1936 mod 111 = 1

Hence 1 + 1 = 2 smile
Re: Number System- 2: Remainders
by Abhinav Chitre - Tuesday, 26 July 2011, 06:00 PM
 

Absolutely Brilliant Article ! Thanks a lot! It covers so many questions asked in CAT or all entrance exams for MBA! smile

Great Work!

Re: Number System- 2: Remainders
by tgdel338 tg - Wednesday, 27 July 2011, 09:05 AM
  Doubt : Frame a digit using only 1 and 0 divisible by 2011.
Re: Number System- 2: Remainders
by TG Team - Wednesday, 27 July 2011, 12:24 PM
 

11111.....2010 digits is divisible by 2011. Now you can put any zeroes after this and number will remain divisible by 2011. smile

Kamal Lohia

Re: Number System- 2: Remainders
by tgdel338 tg - Wednesday, 27 July 2011, 10:33 PM
  @ Kamal Sir : thanks a lot sir but i got stuck in the first step itself...

Why 1111..... 2010 digits are divisible by 2011 ??

Another question : what is the value of n such that
n ! = 3! x 5! x 7 !

Shouldn't the answer be 10 ?

coz

n! = (1x2x3)x (1x2x3x4x5) x (1x2x3x4x5x6x7)
= (1x2x3x4x5x6x7) x (4X2) x (3x3) x (5x2)
=1x2x3x4x5x6x7x8x9x10

Re: Number System- 2: Remainders
by tgdel338 tg - Wednesday, 27 July 2011, 11:07 PM
  Sir please tell me if my logic is right or wrong for question

find the 28383rd term of series: 1234567891011121314....

As explained i have calculated 1*9 = 9 , 90*2=180, 900*3=2700

9 + 180 + 2700 = 2889

Now 28383-2889 = 25494 ( these are all 4 digit numbers)

now 25494/4 = 6373 + 2 that means 2nd digit of 6374

Hence answer is 3
Re: Number System- 2: Remainders
by TG Team - Friday, 29 July 2011, 02:42 PM
 

Hi

Using Euler's Theorem we know that 102010 ≡ 1 mod 2011 as HCF(10, 2011) = 1 and Φ(2011) = 2010.

So 102010 - 1 ≡ 0 mod 2011

and 999...2010 digits ≡ 0 mod 2011

Or 9(111...2010 digits) ≡ 0 mod 2011

as 9 is co-prime with 2011, we can safely conclude that

111...2010 digits ≡ 0 mod 2011.

Kamal Lohia

Re: Number System- 2: Remainders
by tgdel338 tg - Monday, 1 August 2011, 08:34 PM
  Awesome thanks for the explaination sir.
Re: Number System- 2: Remainders
by nishant naveen - Tuesday, 2 August 2011, 01:18 PM
 

i more simpler method:

444^444^444/7

444 mod 7=3, so we now have  3^444^444/7

now take 3^444 => 3^3^148 mod 7

3^3 will give remainder as -1 or 6(consider -1)

now -1^148=1. so now 1^444 mod 7=1.

Re: Number System- 2: Remainders
by Sahej preet Kharbanda - Friday, 5 August 2011, 03:30 PM
 

Hello sir,

im really thankful to u for providing suck agreat material..

but i didnt understand the fosters n wilson's theorem... can u plsss elaborate the theorem for me...

thnks a alot

Re: Number System- 2: Remainders
by TG Team - Friday, 5 August 2011, 05:30 PM
 

Hi Sahej preet Kharbanda

Wilson theorem is simply to find remainders in case of factorials as mentioned by TG above.

(p - 1)! + 1 ≡ 0 mod p where p is a prime number.

=> (p - 1)! ≡ -1 mod p ≡ (p -1) mod p

=> (p - 2)! ≡ 1 mod p.

(N ≡ A mod B means N when divided by B gives remainder A OR A is obtained after decreasing some integral multiples of B from N.)

Kamal Lohia

Re: Number System- 2: Remainders
by Sahej preet Kharbanda - Monday, 8 August 2011, 11:19 PM
 

Hi Kamal

thnks alot for reply!!! but please explain me in simple words... i still didnt get how to find remainders in factorials...

Re: Number System- 2: Remainders
by nisha . - Tuesday, 9 August 2011, 12:27 AM
  hey...can nybdy tell me were to get d solutions for d quize???????

plz help me out!!!!!!!!!

thanx in advance!!!!!!!
Re: Number System- 2: Remainders
by ABHINAV SINHA - Tuesday, 9 August 2011, 11:21 AM
  nice....... cool
Re: Number System- 2: Remainders
by Adhaar khanna - Tuesday, 9 August 2011, 04:36 PM
  literally awesome....
Re: Number System- 2: Remainders
by Adhaar khanna - Tuesday, 9 August 2011, 04:51 PM
  Your answer is correct ,but i guess the no would be 7373. (as the first no of the series will be 1000. so 1000 + 6373th no = 7373). Now, its 2nd digit will be 3. smile
Re: Number System- 2: Remainders
by tgdel338 tg - Sunday, 14 August 2011, 02:36 PM
 

Kamal Sir smile

 

Another doubt on the same lines.

Smallest number consisting only of 0 and 1 and divisible by 225.

Now 10^90/ 225 =1

99999...... 90  nines = 0 mod 225

9(1111......90 ones) = 0 mod 225

Now 9 and 225 are not co prime so what will be the answer ??

How do i proceed for an answer ?

Re: Number System- 2: Remainders
by TG Team - Sunday, 14 August 2011, 03:14 PM
 

Hi

Look, 225 = 5²*9, so last two digits of the number should be 00 as number is to be divisible by 25 and previous 9 digits should be 1 each so that number is also divisible by 9. Hence the required smallest number is: 111 111 111 00. smile

Kamal Lohia

Re: Number System- 2: Remainders
by tgdel338 tg - Monday, 15 August 2011, 03:01 PM
 

Thank you sir but my question is that how do we proceed in questions where two such numbers are not co prime.

In this case 9 and 225.

Thanks for the solution though smile

Re: Number System- 2: Remainders
by pranjal haridas - Tuesday, 16 August 2011, 04:43 PM
 

hi kamal, phi(66) = 20 not 32.

so i think we'll have to do it again.

Re: Number System- 2: Remainders
by TG Team - Wednesday, 17 August 2011, 06:30 PM
 

Hi Pranjal smile

You are correct. Thanks for correction. Now solve the problem. smile

Kamal Lohia

Re: Number System- 2: Remainders
by TG Team - Wednesday, 17 August 2011, 06:33 PM
 

Hi

10 and 225 are not co-prime so your first step altogether is incorrect when you write that 1090/225 = 1.

Check it.

Re: Number System- 2: Remainders
by Himanshu Jaggi - Thursday, 18 August 2011, 11:19 AM
 

Hi TG/Kamal Sir,

Can you please explain how to approach the below question -

a no. N when divided by divisor D leaves remainder 23. when same no. N is divided by 12D remainder is 104. what will be the remainder when the no is divided by 6D

Thanks

Himanshu

Re: Number System- 2: Remainders
by TG Team - Thursday, 18 August 2011, 12:30 PM
 

Hi Himanshu smile

If I start from second statement that N = a*12D + 104, so this is same N which when divided by D gives a  remainder of 23.

i.e. 104 = b*D + 23 or D is a divisor of 104-23 = 81 and also D is greater than 23. So D = 27 or 81.

Thus in any case when same N is divided by 6D, remainder will be given by 104 as a*12D is certainly divisible by 6D. So the required remainder is = 104. smile

Kamal Lohia

Re: Number System- 2: Remainders
by Subhajit Gope - Saturday, 20 August 2011, 05:53 PM
  2^147 / 11
=2^2 * 2^145 / 11
=4 * (2^5)^29 / 11
=4 * (32)^29 / 11
=4 * (-1)^29 / 11
=-4

Remainder = 11-4=7
Re: Number System- 2: Remainders
by Subhajit Gope - Saturday, 20 August 2011, 06:26 PM
  Hi Sid,

20^1000 * 100^1000 / 13
=7^1000 * 9^1000
=(7^2)^500 * (9^2)^500
=(-3)^500 * (3)^500
=3^1000
=3 * 3^999
=3 * (3^3)^333
=3 * (1)^333
=3
Re: Number System- 2: Remainders
by Himanshu Jaggi - Sunday, 21 August 2011, 03:37 PM
  Thanks Sir for the reply.
However, I am not been able to get the below statement -
i.e. 104 = b*D + 23 or D

Cn u pls explain this again?
Re: Number System- 2: Remainders
by destiny unruled - Sunday, 21 August 2011, 08:51 PM
  @ Himanshu

Another way to solve the question is:-
D has to be greater than 23 as remainder on division is 23.
So, 6D > 104
Hence, remainder will be 104

Else,
N = Dx + 23
N = 12Dy + 104
Equating them we will get.
D(x - 12y) = 81
So, D has to be a factor of 81 and greater than 104. So, possible values are 27 and 81

So, 6D is 162 or 324 (both greater than 104)

So, remainder will be 104
Re: Number System- 2: Remainders
by Himanshu Jaggi - Monday, 22 August 2011, 12:53 AM
  Awesum !! Thanks !
Re: Number System- 2: Remainders
by vishnu vardhan - Monday, 22 August 2011, 08:56 PM
  this is very good article of reminders in numbers thanx for itsmile
Re: Number System- 2: Remainders
by In lonely planet i live - Saturday, 17 September 2011, 07:09 PM
  Sir,

Please help me in finding the answer.

Question is : whats the remainder if you will divide 7^0+7^1+7^2+7^3+7^4......+7^365 with 5. Thanks in advance.


Re: Number System- 2: Remainders
by Jayesh Kumar - Saturday, 17 September 2011, 08:39 PM
  Is it 3?
Re: Number System- 2: Remainders
by In lonely planet i live - Saturday, 17 September 2011, 08:54 PM
  but how. can u explain?
Re: Number System- 2: Remainders
by chhavi tripathi - Monday, 12 December 2011, 03:22 AM
  This is super awesome smile smile smile approve thank you smile
Re: Number System- 2: Remainders
by TG Team - Monday, 12 December 2011, 01:56 PM
 

71 ≡ 2 mod5

72 ≡ 4 mod5

73 ≡ 3 mod5

74 ≡ 1 mod5

and after that pattern of remainders start repeating. Also sum of the remainders obtained by these four numbers is 0.

So the final remainder will be 70 + (71 + 72 + 73 + 74) + (75 + 76 + 77 + 78) + ... + (7361 + 7362 + 7363 + 7364) + 7361 ≡ 1 + (2 + 4 + 3 + 1) + (2 + 4 + 3 + 1) + ... + (2 + 4 + 3 + 1) + 2 mod5 ≡ 1 + (0) + (0) + ... + (0) + 2 mod5 ≡ 3 mod5. smile

Kamal Lohia

Re: Number System- 2: Remainders
by Ani Rai - Monday, 12 December 2011, 04:49 PM
 

hi.. help me in dis questions..

what is the remainder of : (3^5^7^9)/41

 

 

Re: Number System- 2: Remainders
by TG Team - Monday, 12 December 2011, 05:25 PM
 

Hi Anil smile

34 = 81 ≡ -1 mod41

=> 38 ≡ 1 mod41

=> 38k ≡ 1 mod41

So we need to find the remainder when 57^9 is divided by 8.

Now we know that 52 ≡ 1 mod8, so 52k ≡ 1 mod8.

And 79 ≡ 1 mod2 = 2k + 1.

So going reverse we can say that 57^9 ≡ 52k+1 ≡ 5 mod8 = 8k + 5.

Hence the given number can be re-written as: 38k+5 ≡ 35 mod41 ≡ -3 mod41 ≡ 38 mod41. smile

Kamal Lohia 

 

Re: Number System- 2: Remainders
by tgdel574 tg - Friday, 13 January 2012, 09:23 PM
  hie AMI13

(3^5^7^9)/41
we know that 3 n 41 are co prime ..means prome to each other..

by using euler's theorm v cn say ..3^40/41 gives remainder 1 ....now we have to make 5^7^9 a multiple of 40k

n we know that when divided by 40...5^1 gives remainder 05 ...
5^2 25
5^3 05
5^4 25

which means odd power gives 5 n even power 25....kk

so now 5^7(odd) ..& if power is odd then all other power remains odd..which means it will give remainder 5 when divided by 40...

so we can write in this way..3^40k+5/41..

n v know that 3^40k/41 gives remainder 1.& 3^5/41 gives remainder 38.....

so now the remainder is 1*38 =38..final ans
Re: Number System- 2: Remainders
by pavani tk - Sunday, 22 January 2012, 09:21 PM
  What is the remainder when 6^83 + 8^83 is divided by 49 ??

pie(49)=42
6^(84-1)+8^(84-1)
so when (6^83+8^83)%49 = -6-8 = -14 =35(i.e;49-14)

"a^pie(N)%N = 1"
Re: Number System- 2: Remainders
by shalin saxena - Saturday, 4 February 2012, 02:04 PM
  yes 47 is the answer
to the que 3^101/77
Re: Number System- 2: Remainders
by shalin saxena - Saturday, 4 February 2012, 02:33 PM
  a very nice article sir
can u tell me the ans of
32^32^32 remainder when divided by 3,5,7,9,6
Re: Number System- 2: Remainders
by Radha Chawla - Monday, 12 March 2012, 08:15 PM
 

Hi Kamal Sir,

I too have similar doubt.

How 36= cD+7

Pl clarify..

 

Thanks,

Reeti

Re: Number System- 2: Remainders
by Radha Chawla - Monday, 12 March 2012, 08:39 PM
 

Hi,

I think i got the solution to it: 9aD+36=9N

also, bD+7=9N

Therefore, 9aD-bD+36-7=0

Let, 9aD-bD=cD

so, 36=cD+7

cd=29

Pl, correct me if i m wrong!

 

Thanks,

Reeti.

 

Re: Number System- 2: Remainders
by TG Team - Wednesday, 14 March 2012, 12:23 PM
 

No Reeti/Radha smile

There is no need to correct you. You are absolutely right. smile

Kamal Lohia

Re: Number System- 2: Remainders
by achal singhal - Monday, 26 March 2012, 03:08 PM
 

Hi TG,

Your articles are just too good. The way you present the concepts is just amazing. Could you plz write also on RATIO & PROPORTION! That would be of great help for me.

Thanks 

Achal

Re: Number System- 2: Remainders
by In lonely planet i live - Wednesday, 18 April 2012, 12:05 AM
  Really great post sir.

one doubt:
the remainder of a number when divided by a divisor Q is 9. the remainder when 4 times the number is divided by Q is 17. find Q?

thanks in advance.
Re: Number System- 2: Remainders
by arsh arora - Wednesday, 18 April 2012, 01:44 AM
 

hi lonely planet,

  Q is 19 here..28/19= rem 9,  112/19=rem 17

Re: Number System- 2: Remainders
by In lonely planet i live - Wednesday, 18 April 2012, 04:18 AM
  Can you please explain, how u got to the answer?
Re: Number System- 2: Remainders
by arsh arora - Wednesday, 18 April 2012, 04:25 AM
  use hit and trial....start from 10 as Q...u will stop at 19!!!!...take N corresponding to the Q!!..hope u got it!!
Re: Number System- 2: Remainders
by In lonely planet i live - Wednesday, 18 April 2012, 04:35 AM
  sorry brother, lets avoid hit and trial method here. First we will try to solve it properly otherwise hit and trial will be messy some times sometimes. anyways thanks for the attempt.

(it's always good to have hit and trial method but not mandatory)
Re: Number System- 2: Remainders
by arsh arora - Wednesday, 18 April 2012, 04:42 AM
  alrght man!!..but i think in dis case if we will go by the normal approach, time wud be consumed much!!
Re: Number System- 2: Remainders
by TG Team - Wednesday, 18 April 2012, 09:57 AM
 

Hi smile

There is no need to try a wague hit. Just be logical.

See if N is the number and Q is the divisor, as you mentioned then we can write:

N = aQ + 9 as N gives a remainder of 9 when divided by Q. That means Q > 9.

Also it if we multiply above equation with 4, we get that:

4N = 4aQ + 36. But it is given that 4N gives a remainder of 17 when divided by Q, that means 36 when divided by Q gives a remainder of 17.

Or in other words, 36 = bQ + 17

i.e. 36 - 17 = 19 = bQ

i.e. Q is a divisor of 19.

Combining both conclusions about Q, we get to know that Q is a divisor of 19 greater than 9. Certainly Q is 19 only. smile

Kamal Lohia 

Re: Number System- 2: Remainders
by arsh arora - Wednesday, 18 April 2012, 10:15 AM
  thank you sir!!!!...
Re: Number System- 2: Remainders
by In lonely planet i live - Wednesday, 18 April 2012, 01:22 PM
  gr8!!!!

Thanks.
Re: Number System- 2: Remainders
by Mohit Sharma - Wednesday, 18 April 2012, 06:49 PM
  Thank u sir... gr8 approach...smile
Re: Number System- 2: Remainders
by Prerna Golani - Saturday, 12 May 2012, 02:34 PM
  Please Solve the following question

ABCABC is a six digit number(therefore both As,both Bs and both Cs are same).C is even digit

1.Is ABCABC divisible by 14
a.Yes b.No c.divisible if A is even d.divisible if B is even e.None of These

2.Is ABCABC divisible by 13
a.Yes b.No c.divisible if A=2 d.divisible if B=5 e.None of These
Re: Number System- 2: Remainders
by TG Team - Monday, 14 May 2012, 12:40 PM
 

Hi Prerna smile

See ABCABC = ABC × 1001 = ABC × 7 × 11 × 13

As C is even digit i.e. ABC, the three digit number is even and divisible by 2.

1. Yes, the number is certainly divisible by 14. (a)

2. Yes, the number is certainly divisible by 13. (a)

Kamal Lohia

Re: Number System- 2: Remainders
by Radha Chawla - Wednesday, 23 May 2012, 02:51 PM
 

Thanks a lot sir!!!

 

Regards,

Radha

 

Re: Number System- 2: Remainders
by Radha Chawla - Wednesday, 23 May 2012, 03:03 PM
 

Sir,

Please explain the solution to the below Q.)

A = {179, 180, 181,…..,360}. B is a subset of A such that sum of no two elements of B is divisible by 9. The number of elements in B cannot exceed????

a)102    81)   c)101   d)82   e)  None of these.

 

Thanks,

Radha

Re: Number System- 2: Remainders
by Mohit Sharma - Thursday, 24 May 2012, 07:03 AM
  Hi radha,
Ans has to be none of these. As max value of elements is 121.
I hope the ans is correct. Let me knw.

Thanks n Regards,
Re: Number System- 2: Remainders
by Praveen Patel - Thursday, 19 July 2012, 05:29 PM
 

The expansion got a fault

1^39+2^39+3^39+....+12^39     !=    (1+2+3+...+12)(1^38+2^38+....12^38)

 

Try for some small no.you will get it.

Re: Number System- 2: Remainders
by rohith kotagiri - Thursday, 27 September 2012, 09:02 AM
  Dear TG,

Help me solve the following., thoughtful
Find the reminder when (10^10)+(10^100)+(10^1000)+.....+(10^10000000000) is divided by 7.
Re: Number System- 2: Remainders
by gaurav midha - Saturday, 29 September 2012, 07:10 PM
  Hi Rohit,

http://totalgadha.com/mod/forum/discuss.php?d=1369

It is explained in the above lesson.



Regards,
GM
Re: Number System- 2: Remainders
by shobana raut - Monday, 1 October 2012, 07:08 PM
  can some one help me solving this prob
128^1000/153
Re: Number System- 2: Remainders
by amresh kumar - Tuesday, 2 October 2012, 12:23 AM
  Getting 52 !!

E(153) = 96

2^7000 Mod 153

2^88 Mod 153

(1024)^4 * (1024)^4 * 256 Mod 153

(106)^4 * (106)^4 103 Mod 153

52 * 52 * 103 Mod 153

52 Mod 153

PS: My first ever post on TG
Re: Number System- 2: Remainders
by shobana raut - Tuesday, 2 October 2012, 11:06 AM
  thanks for rply amresh....bt could you explain this step

(106)^4 became 52 in nxt step did you do it by normal method of multiplying or any short cut????
Re: Number System- 2: Remainders
by Panna Lal Patodia - Saturday, 16 February 2013, 06:21 PM
  It is very simple provide you know Euler's Generalization of Fermat's Little Theorem and Chinese Remainder Theorem. If you do not know about these, either read some decent book on "Theory of Numbers" or simple Google it. Now, the answer part:
3^101 (mod 77). As 77 = 7x11. We calculate it in two parts.

3^101 (mod 7) = 3^5 (mod 7) because 101 is divided by 6 gives remainder of 5.
= 5 (mod 7).

3^101(mod 11) = 3^1 (mod 11) because 101 divided by 10 gives remainder of 1.
= 3 (mod 11).

Now, using Chinese Remainder Theorem, we get 47 (mod 77).

So, the answer is 47.

Regards,

P L Patodia

Re: Number System- 2: Remainders
by Ananda Dutta - Monday, 8 April 2013, 04:23 PM
  A = Remainder([(10^10)+(10^100)+(10^1000)+....+(10^10000000000)]/7)
Remainder(10/7) is 3
A = Remainder([3^(10^1)]+[3^(10^2)]+[3^(10^3)]+[3^(10^4)]+[3^(10^5)]+[3^(10^6)]+[3^(10^7)]+[3^(10^8)]+[3^(10^9)]+[3^(10^10)]/7)
Now we know that as 7 is an prime number,
So, 3^(7-1)/7 = 3^6/7 = 1
(3^6/7)^n = 1 (For any integral value of n)
A =Remainder([(3^6/7)^n x 3^4]+[(3^6/7)^n x 3^4]+[(3^6/7)^n x 3^4]+[(3^6/7)^n x 3^4]+[(3^6/7)^n x 3^4]+[(3^6/7)^n x 3^4]+[(3^6/7)^n x 3^4]+[(3^6/7)^n x 3^4]+[(3^6/7)^n x 3^4]+[(3^6/7)^n x 3^4]/7)
A =Remainder([3^4]+[3^4]+[3^4]+[3^4]+[3^4]+[3^4]+[3^4]+[3^4]+[3^4]+[3^4]/7)
A =Remainder(10 x 3^4 / 7)
Now (3^3/7) = Remainder (-1/7)
=Remainder (10x3x(-1)/7)
=Remainder (-30/7)
=Remainder (-2/7)
=Remainder (5/7)
Ans. 5
Re: Number System- 2: Remainders
by anik gupta - Tuesday, 23 April 2013, 07:11 PM
  (10^10)/7=3*(((3)^3)^3)/7=(27^3)*3/7=((28-1)^3)*3=-3/7
similarly other 9
and we get
-30/7=5
Re: Number System- 2: Remainders
by anila . - Sunday, 12 May 2013, 02:59 PM
  ques Determine number of terms in expansion of (1+a1+a2+a3)^8????
Re: Number System- 2: Remainders
by DeeScript .in - Friday, 7 June 2013, 11:06 AM
  hi abhinav......there is mistake.....
4^4=64.........????????????4^3 is 64
 i.e (-4^3)148 now the answer will be 1.

Re: Number System- 2: Remainders
by Raunak Kumar - Wednesday, 17 July 2013, 09:26 AM
  1st of all segregate the terms of 444^444^444/7 as
444^444/7. from here we will get the remainder as 1,
now 444^1/7=3 ans.
Re: Number System- 2: Remainders
by Raunak Kumar - Wednesday, 17 July 2013, 09:48 AM
  56 is the answer
Re: Number System- 2: Remainders
by Raunak Kumar - Wednesday, 17 July 2013, 09:55 AM
  on 3^18/77 we are getting remainder as -1. 3^(18*5+11)
(-1)^18*5. 3^11/77
3^11/77 the remainder is 21
therefore, -21 is the final remainder as -1^ odd is -1,.
77-21=56
Re: Number System- 2: Remainders
by saumya Gupta - Wednesday, 31 July 2013, 10:15 PM
  What is the remainder of 59^73^5! when divided by 37?
Re: Number System- 2: Remainders
by navya VELURI - Tuesday, 6 August 2013, 09:10 AM
  can anyone please repost number system -1 forum.. i really need it plzzz sad sad
Re: Number System- 2: Remainders
by shivam mehra - Monday, 12 August 2013, 10:47 PM
  E(37)=36


59^73===59^36k+1 = 59^36k x 59 = 1x22=22
Re: Number System- 2: Remainders
by madhavi kumari - Monday, 3 March 2014, 01:17 PM
  hello,

This is nice article , it helped in knowing remainder tricks well.
But i m getting one doubt, please seek to solve it.
To find the remainder for 617 + 1176 when divided by 7.
I m getting the remainder as 1 , but the correct answer is 0.
please explain me how it is .

thank u !
Re: Number System- 2: Remainders
by TG Team - Monday, 3 March 2014, 01:46 PM
 

Hi Madhavi smile

Certainly 617 + 1176 = 1793 = 1 mod 7.

But I am sure that you have noted the question wrongly.

IMO, it is to find the remainder when 617 + 176 is divided by 7. And that is zero. smile

Kamal Lohia 

Re: Number System- 2: Remainders
by TG Team - Monday, 3 March 2014, 03:30 PM
 

Hi Anila smile

I know that it's bit late but beter late than ever. smile

Number of terms in the expansion (1 + a1 + a2 + a3)8 is simply number of whole number solutions of x + y + z + w = 8 which is equal to C(8+4-1, 4-1) = C(11, 3) = 165. smile

Kamal Lohia

Re: Number System- 2: Remainders
by TG Team - Monday, 3 March 2014, 04:50 PM
 

Hi Radha smile

In the given set of numbers A, we have 9 types of numbers, based on their remainders with 9.

9k+8 - 21 numbers

9k - 21 numbers

9k+1 - 20 numbers

9k+2 - 20 numbers

9k+3 - 20 numbers

9k+4 - 20 numbers

9k+5 - 20 numbers

9k+6 - 20 numbers

9k+7 - 20 numbers

If no two numbers of subset of A i.e. set B add upto a multiple of 9, then we can either take numbers of the form 9k+1 OR 9k+8 but not both. Similarly either 9k+2 OR 9k+7 and so on. Also we can take atmost one number of the form 9k.

So considering the forms with more numbers we get the maximum possible number of elements in set B as = 1 + 21 + 20 + 20 + 20 = 82. (d) smile

Kamal Lohia

Re: Number System- 2: Remainders
by Anirudh Singh - Tuesday, 4 March 2014, 12:49 PM
  (6^17+17^6)/7=?
Re: Number System- 2: Remainders
by TG Team - Tuesday, 4 March 2014, 01:00 PM
 

Hi Anirudh smile

Deal them independently. That'll be easier, I guess.

617 ≡ (-1)17 mod 7 ≡ -1 mod 7

176 ≡ 36 mod 7 = 272 mod 7 ≡ (-1)2 mod 7 ≡ 1 mod 7.

Now combining both, it is easy to see that 617 + 176 ≡ -1 + 1 = 0 mod 7. smile

Please go through above lesson and previous post/comments thoroughly. I am sure you will easily get it.

Kamal Lohia  

Re: Number System- 2: Remainders
by Abheek Das - Saturday, 19 July 2014, 01:55 PM
  I wanted to find out the best method of solving questions of the form :

(7^26 x 5^83 )/100?

My Solution which didn't get me anywhere but still

The prime factorisation of 100 =2^2 x 5^2
So under 7^26 i wrote 5^2 and under 5^83 i wrote 2^2.

Next step : 5^83=(4+1)^83, therefore when i divide this by 2^2 i get remainder 1.

Now using Euler Totent Function Number of number co-prime to 5^2 are 20.
So 7^26 is written as 7^20 x 7^6.
7^20 divided by 5^2 leaves remainder 1 as the power is a multiple of the number of number co-prime to 5^2.ie 20
7^6 can be expressed as 7^2x 7^2 x 7^2.
7^2 is 49, therefore on division by 5^2 i get remainder 24 or -1.
Thus -1 multiplied odd number of times leaves -1 only and that divided by 25 gives me 24.
Therefore my final answer is 24.
I don't have the solution key but all the options are multiples of 5 i.e..
50,25,35,45,75
Re: Number System- 2: Remainders
by Amit Agarwal - Saturday, 19 July 2014, 07:43 PM
  Better find the last two digit of the individual number and multiply it...Is the answer 25?
Re: Number System- 2: Remainders
by Sangram Pattnaik - Friday, 25 July 2014, 12:39 AM
  N = 33
D = 29

R[N/D] = 4

R[9N/D] = 9*R[N/D] = 9*4 = 36
Now, R[36/D] = 7
So, D = 26 & N = 33
Re: Number System- 2: Remainders
by Sangram Pattnaik - Friday, 25 July 2014, 01:21 AM
  53
Re: Number System- 2: Remainders
by Abheek Das - Saturday, 2 August 2014, 11:07 PM
  An interesting question

16! Divided by 19.Find remainder ?


Hint
(P-2)!/P = 1 (remainder)

How many of N1,N2,N3 are prime?
N1=11111.........111(91 times)
N2=11111.......111(243 times)
N3=11111.......111(57 times)

Please remember something about solving the question given above.
6k+1 is not a method of checking if a number is prime or not.

Re: Number System- 2: Remainders
by Abheek Das - Monday, 4 August 2014, 10:04 PM
  I somehow obsess over the nitty-gritty's of number system.
And hence the following question present's itself.

QUESTION.

What is the remainder when 57^99+55^99 is divided by 224?

My Approach
a^n+b^n is always divisible by a+b if n is odd
Thus the above expression is also divisible by 112.(always)

The sum of the terms in the numerator would necessarily leave an even number( odd+odd = even)

Therefore, the numerator is divisible by both 112 & 2
Hence should be divisible by the product 112 x 2=224

The remainder should come out to be 0.

But the official answer is 112.

My question is in two-parts?
A. Similar question which was 74^13-41^13+75^13-42^13 is divisible by 66 ( following the logic i put forth in the above mentioned lines)

What is the difference in this question and the one i posted above ?

Regards,
Abheek
Re: Number System- 2: Remainders
by TG Team - Tuesday, 5 August 2014, 12:57 PM
  Hi Abheek smile

You answer me, if a number is divisible by 112 and 2 as well, then will the number be certainly divisible by 112*2 = 224 too?

As a hint, you can check with example of the smaller numbers which satisfy both the above conditions.

Actually if a number is divisible by a as well as b, then the number is divisible by LCM(a, b) and not a*b. It'll appear as a*b if HCF(a, b) = 1 as in the similar question mentioned by you.

Hope it clears your doubt. smile

Kamal Lohia
Re: Number System- 2: Remainders
by Abheek Das - Tuesday, 19 August 2014, 10:56 AM
  Let d be the greatest common divisor of 2^{30^{10}}-2 and 2^{30^{45}}-2. Find the remainder when d is divided by 2013.
I found this question on the princeton math forum.
Please give it a thought.
Re: Number System- 2: Remainders
by shalvi sharan - Friday, 3 October 2014, 01:40 PM
  Hi. Could anyone help me in solving this question.
3^101/77
Re: Number System- 2: Remainders
by Vishal yadav - Saturday, 14 March 2015, 06:46 PM
  hey,how u took common
,huh.
Re: Number System- 2: Remainders
by Udit D - Wednesday, 22 April 2015, 08:08 PM
  How 17^36 = 1 (mod37) please explain
Re: Number System- 2: Remainders
by Devashish Acharya - Sunday, 30 August 2015, 07:33 PM
  Ans is 1

Solution:
By Fermat's Little Theoram:

a^(p-1) - 1 = kp ( where k is a constant & p is prime number)
and a & p co-prime.

a = 17, p = 37

17^(37-1) - 1 = 37k
=> 17^36 = 37k + 1

Hence, Remainder = 1