New Batches at TathaGat Delhi & Noida!               Directions to CP centre
Divisibility
by Total Gadha - Saturday, 10 February 2007, 02:22 AM
 

1.6 A Divisibility by 2, 4, 8, 16, 32..

A number is divisible by 2, 4, 8, 16, 32,.. 2 n when the number formed by the last one, two, three, four, five...n digits is divisible by 2, 4, 8, 16, 32,..2 n respectively.

Example: 1246384 is divisible by 8 because the number formed by the last three digits i.e. 384 is divisible by 8. The number 89764 is divisible by 4 because the number formed by the last two digits, 64 is divisible by 4.

1.6 B Divisibility by 3 and 9

A number is divisible by 3 or 9 when the sum of the digits of the number is divisible by 3 or 9 respectively.

Example: 313644 is divisible by 3 because the sum of the digits- 3 + 1 + 3 + 6 + 4 + 4 = 21 is divisible by 3.

The number 212364 is divisible by 9 because the sum of the digit- 2 + 1 + 2 + 3 + 6 + 4 = 18 is divisible by 9.

1.6 c Divisibily by 6, 12, 14, 15, 18..

Whenever we have to check the divisibility of a number N by a composite number C, the number N should be divisible by all the prime factors (the highest power of every prime factor) present in C .

divisibility by 6: the number should be divisible by both 2 and 3.

divisibility by 12: the number should be divisible by both 3 and 4.

divisibility by 14: the number should be divisible by both 2 and 7.

divisibility by 15: the number should be divisible by both 3 and 5.

divisibility by 18: the number should be divisible by both 2 and 9.

EXAMPLES

35. The six-digit number 73A998 is divisible by 6. How many values of A are possible?

Answer: Since the number is ending in an even digit, the number is divisible by 2. To find divisibility by 3, we need to consider sum of the digits of the number.

The sum of the digits = 7 + 3 + A + 9 + 9 + 8 = 36 + A.

For the number to be divisible by 3, the sum of the digits should be divisible by 3. Hence A can take values equal to 0, 3, 6, and 9.

Answer = 4

1.6 d Divisibility by 7, 11, and 13

Let a number be ....kjlhgfedcba where a, b, c, d, are respectively units digits, tens digits, hundreds digits, thousands digits and so on. Starting from right to left, we make groups of three digit numbers successively and continue till the end. It is not necessary that the leftmost group has three digits.

Grouping of the above number in groups of three, from right to left, is done in the following manner à kj,ihg,fed,cba

We add the alternate groups (1 st , 3 rd , 5 th etc.. and 2 nd , 4 th , 6 th , etc..) to obtain two sets of numbers, N 1 and N 2 .

In the above example, N 1 = cba + ihg and N 2 = fed + kj

Let D be difference of two numbers, N 1 and N 2 i.e. D = N 1 – N 2 .

à If D is divisible by 7, then the original number is divisible by 7.

à If D is divisible by 11, then the original number is divisible by 11

à If D is divisible by 13 then the original number is divisible by 13.

Corollary:

Any six-digit, or twelve-digit, or eighteen-digit, or any such number with number of digits equal to multiple of 6, is divisible by EACH of 7, 11 and 13 if all of its digits are same .

For example 666666, 888888888888 etc. are all divisible by 7, 11, and 13.

Example

36. Find if the number 29088276 is divisible by 7.

Answer: We make the groups of three as said above- 29,088,276

N 1 = 29 + 276 = 305 and N 2 = 88

D = N 1 – N 2 = 305 – 88 = 217. We can see that D is divisible by 7. Hence, the original number is divisible by 7.

37. Find the digit A if the number 888…888A999…999 is divisible by 7, where both the digits 8 and 9 are 50 in number.

Answer: We know that 888888 and 999999 will be divisible by 7. Hence 8 written 48 times in a row and 9 written 48 times in a row will be divisible by 7. Hence we need to find the value of A for which the number 88A99 is divisible by 7. By trial we can find A is = 5.

Answer = 5.

Re: Divisibility
by bimal mohan - Friday, 1 June 2007, 05:45 AM
 

Dear  TG  sir,

   really   a  fine  article.....very  smooth  and   to  the  point...thnx   for  that.....

Here   i   wud  like  to   quote   a  problem   from  your   site  :

*  [ 43  pow444 + 34 pow333]  is   divisible  by:    5  /  2  /9  /11  ??

     here   43  pow 444 ==>  last  digit =1

     and  34   pow  333==>  last  digit =4

so  overall,  last  digit  =5  ==>   divisible  by 5 .

but   i   could  not  confirm  whether  it   is  divisible by  9  /11   or  not .....

what's   wud  be  the  method  ??

 

Re: Divisibility
by rishi sawla - Saturday, 7 July 2007, 06:08 PM
 

 use this theorem

An + Bn is divisible by A+B when n is odd

u can easily convert above powers 444 and 333 to 111 which is odd.

then u can get A+B, now if A+B is divisible by 9/11 then the whole

number will also be divisible.

however this is not the best method ,but this is what i tried.

Re: Divisibility
by rmmozhi prathiba - Monday, 16 July 2007, 05:10 PM
 

Hi TG how to solve this problem??

10000! = (100!)K × P, where P and K are integers. What can be the maximum value of K?

Re: Divisibility
by DEEPAK KANSAL - Tuesday, 24 July 2007, 05:11 PM
 

43 ^ 444 = (45-2)^444

all terms will be divisible by 9 except 2^444.

2^444 = (9-1)^148

remainder is 1

similarly for 34^333 = (36-2)^333

all the terms will be divisible by 9 except 2^333

2^333 = (9-1)^111

remainder is -1 or 8

if we add both the remainders thus 8+1 = 9

hence divisible by 9

Re: Divisibility
by DEEPAK KANSAL - Tuesday, 24 July 2007, 05:11 PM
 

43 ^ 444 = (45-2)^444

all terms will be divisible by 9 except 2^444.

2^444 = (9-1)^148

remainder is 1

similarly for 34^333 = (36-2)^333

all the terms will be divisible by 9 except 2^333

2^333 = (9-1)^111

remainder is -1 or 8

if we add both the remainders thus 8+1 = 9

hence divisible by 9

Re: Divisibility
by Mayur Mantri - Friday, 27 July 2007, 04:11 PM
 

Hi Pratibha

In the above problem first check  the number of zeroes in 1000! which is 2499.Then find number of zeroes in 100! which is 24.(To calculate this,check the note in bold at the end). 

Now, to get  as many zeroes as possible closest to and less than equal to 2499 we can multiply 24 with 104.(As any number followed by x zeroes raised to power n give x*n zeroes in the final product).

Hence maximum value of k should be 104 as 104*24 = 2496.

NjoYY

Highest power of x which divides n! = [n/x] +[n/x2] +[n/x3] + ...so on.

Where [ ] is the greatest integer function.

Re: Divisibility
by Total Gadha - Saturday, 28 July 2007, 04:06 AM
  Hi Mayur,

The value of k can be found by seeing the highest power of 97 (highest prime number in 100!) and the highest power of 297 (the highest power of 2 in 100!) in 10000!. Whichever is the lowest gives us the answer.

Total Gadha
Re: Divisibility
by tripti pathak - Monday, 30 July 2007, 09:26 AM
  hI TG,

In case of 17017 wen we apply the rule to check for divisibility by 7 ,11
i get two set 170 ,17
 170 -17=153 which is not divisible by 7
but the number 17017 is divisible of 7
2431*7 =17017
Re: Divisibility
by Catapult!!! On the way - Monday, 30 July 2007, 09:33 AM
  hi Tripti..
17017
U have to make groups frm right 017 and 17 diff is 0 and hence divisible by 7...smilesmilesmile
Re: Divisibility
by DEEPAK KANSAL - Monday, 30 July 2007, 04:41 PM
 

hi TG

please explain the logic

Re: Divisibility
by anil kalwakuntla - Tuesday, 31 July 2007, 02:17 AM
  hi mayur can u pls elobrate ur ans
Re: Divisibility
by Total Gadha - Tuesday, 31 July 2007, 03:39 AM
  Hi Tripti,

You make groups of three from right to left. In case of 17017, you get groups 17 and 017. The difference = 17 - 17 = 0, hence the number is divisible by 7 (and also by 11 and 13)

Total Gadha
Re: Divisibility
by tripti pathak - Tuesday, 31 July 2007, 01:03 PM
  Thanks TGapprove
Re: Divisibility
by harshhal bandekar - Tuesday, 14 August 2007, 11:57 PM
 

TG,Plz help me out with these problem  on number system

Let N=123456789123456789…..up to 180 digits. Find the remainder when N is divided by11.

1. 2  

2. 3

3. 4

4. 0                            

 

Which of the following is the least possible natural no. n such that n! is divisible by 1089

  1. 11
  2. 44
  3. 33
  4. 22

 

Re: Divisibility
by arsenal . - Saturday, 25 August 2007, 09:01 PM
 

N is made of 10 repeating sets where each set is

1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9

-   -   -   -   -   -   -   -   - 

By rule of divisbility for 11, add nos which are underlined into one and the remaining in to another

Both the totals are 45. Difference b/n two = 0

Multiply this for 10 sets. Net diff = 0

Hence the no is divisible by 11.

 

TG sir, can u confirm the approach.

 

 

Re: Divisibility
by arsenal . - Saturday, 25 August 2007, 09:08 PM
 

1089 = 3^2 * 11^2

Take the highest prime factor.

The power of 11 is 2. So by using the highest power in a factorial concept

the answer should be 22.

 

Could you please confirm if the ans is correct

Re: Divisibility
by Saikat Chakraborty - Sunday, 26 August 2007, 10:07 AM
 

thanks TG for that helpful article.

to check the divisibility by the prime nos 7, 11, 13 was really helpful.

in the same way can we get some more generalized one that can be applied to all the primes ?

Re: Divisibility
by Priyanka Ghosh - Monday, 27 August 2007, 12:04 PM
 

Hi Vamsi,

I think your approch is right.Even I thought of teh same approach to solve it... smile

 

 

Re: Divisibility
by writtika maitra - Thursday, 13 March 2008, 11:25 AM
 

i didn't understand the soln to d prob 10000!=(100!)^k*p

can anyone explain it to me.

Re: Divisibility
by Sumit Gite - Sunday, 16 March 2008, 02:26 AM
 

Hi Bimal,

This is not divisible by 9. I have got the method....and I have verified it with scientific Calc(PC) . Just check there is a remainder of 2.

Before I tell the method I want to verify it with u people if I m correct.

ques !!!!
by Chitrang Dalal - Monday, 23 June 2008, 03:46 PM
 

@TG

Help me wid this plzz!!

Find the remainder when N =1 × 2 + 2 × 3 + 3 × 4 + ... + 98 × 99 + 99 × 100 is divided by 101.
................
 wat i did is
N =1 × 2 + 2 × 3 .........+ -2×-3 + -2 × -1
    (used the concept of -ve remainder)
   = 2 (1 × 2 + 2 × 3 + 3 × 4 +...... + 49 × 50) 
now this is not divisible by 101
 
if i do direct summation of N.....then the number is divisible by 101
wats the problem with the first approach 
 
anxiously xpecting an xplanation....thanks in advnace
Re: Divisibility
by Biswajit Sen - Monday, 23 June 2008, 08:06 PM
 

the answer is 33....

1089=33*33

33! = 33*32*31*.......11*10*....3....

      = 33*11*3* some value

      = 33*33*some value

therefore 33! / 33*33 = some value

therefore the least value of n is 33.

please correct me if i am wrong.smile

Re: Divisibility
by shiva boddepalli - Thursday, 26 June 2008, 05:01 PM
 

hi sir,i have a better method of finding the divisibility of a number by 7,13,17,19.the method is known as osculator method.first we have to know the concept of osculator.the osculator of 7 is 2 ( 7* 3 = 21 = 20+1 here we are adding 1 so we need to consider the osculator as negative osculator )

 similarly for 13 ,13 * 3 = 39 = 40-1 so the osculator is 40 and is one more osculator and the value is 4 and for 17 ,17 * 3 = 51 = 50+1 so again the osculator is negative and it is 5 and for 19 it is one more osculator  19 = 20-1 ,so the osculator is 2 .

if you didn't understand the concept of osculator just remember these

for 7 osculator = 2 , sign = ' - '

for 13 osculator = 4 , sign = ' + '

for 17 osculator = 5 , sign = ' - '

for 19 osculator = 2 , sign = ' + '

lets proceed with an example

55277838 is considered and to check it's divisibility by 7.

5527783 8  : 5527783 - 8 *2 =5527767 ( here we have taken the units digit and multiplied it by osculator 2 and subtracted it from the remaining part as the sign is negative as suggested above we have subtracted )

the next step is the following is repeated for the resulting value in the above step

552776   : 552776 - 7 *2 = 552762

55276 2 : 55276 - 2 * 2 = 55272

5527 2 : 5527 - 2 *2 = 5523

552 3 : 552 - 3 *2 = 546

54 6 : 54 - 6*2 = 42 as 42 is divisible by 7 so the number is divisible by 7.

Re: Divisibility
by Total Gadha - Thursday, 26 June 2008, 10:44 PM
 

Hi Shiva,

Have already discussed this before:

http://totalgadha.com/mod/forum/discuss.php?d=84

 

Total Gadha

Re: Divisibility Rule for 7, 11, and 13
by Dhiraj Kain - Tuesday, 1 July 2008, 12:15 AM
 

Hi TG

I think that the divisibility rule for divisibility 7, 11, and 13 is wrong.

If we consider a number 1414, which is divisible by 7. The rule fails here.

As per rule,

(1)(414) - grouping of 3 elements from the right.

1 - 9     - summation of alternate groups

8          - number not divisble by 7

So, rule fails here. Same is the case for number 1313, which is divisible by 13.

Please help me, if my understanding to the rule is wrong...

 

Cheers

Re: Divisibility Rule for 7, 11, and 13
by Total Gadha - Tuesday, 1 July 2008, 02:30 AM
  1, 414 ---> 414 - 1 = 413 (divisible by 7)

1, 313 ---> 313 - 1 = 312 (divisible by 13)
Re: Divisibility
by Satyam Gadha - Tuesday, 1 July 2008, 07:54 PM
  Che TG,
Can you please elaborate more on this approach. I know this method but I am not able to grasp it in one go. A few more sentences here and there will be a great help.

Thanks
Satyamgadha
Re: Divisibility
by shubham singh - Tuesday, 1 July 2008, 08:30 PM
  hi biswajit.we have to find out the least value of n and the least value will be 22. as 1089=(3^2)*(11^2).so n in n! should be 22
Re: ques !!!!
by Satyam Gadha - Wednesday, 2 July 2008, 01:05 PM
  Che Dalal,

Your first approach is wrong. I will explain it with a simple example.

When you divide 1 x 2 by 101, you get a remainder = 2
But when you divide 99 x 100 by 101, you can not get remainder as = -2 x -1
Because, you can divide only one number in this multiplication, i.e either 99 or 100 and not both. If it had been addition (+), your approach would have been correct.
Hope this clears the cloud.

Thanks
SatyamGadha
Re: Divisibility
by Satyam Gadha - Wednesday, 2 July 2008, 03:06 PM
  Che TG,

The value of k can be found by seeing the highest power of 97 (highest prime number in 100!) and the highest power of 297 (the highest power of 2 in 100!) in 10000!. Whichever is the lowest gives us the answer.

The first part of the above basic was known to me but I never went that deep to conceptualize the second one.
However, I would try to explain  it to my buddies here.
Consider this formula
Highest power of x in n! = [n/x] + [n/x2] +[n/x3]+....
where [] is the greatest integer function.
For e.g Find the number of zeroes in 127!
Sol: 10 =  21 x 51
  So number of zeroes will be calculated using the lowest of following:
  Highest power of 51 in 127! = [127/5] + [127/52] + [127/53]
                                         = 25 + 5 + 1 = 31
  Highest power of 21 in 127! = [127/2] + [127/22] + [127/23] + ....+ [127/26]
                                         =  63 + 31 + 15 + 7 + 3 + 1 =  120
  So the number of zeroes are 31.

Since 100!= 100x99x98x97x96x......x3x2x1
Similarly, here in this problem we have 971 (the highest prime number in 100!)  instead of 51.     
Highest power of 97 in 10000! = [10000/971] + [10000/972]
                                         = 103 + 1 = 104

Now, lets see why TG has mentioned the the highest power of 297 (the highest power of 2 in 100!) in 10000!.
since the highest power of 2 in 100! = [100/2] + [100/22] + [100/23] + [100/24]
                                                     = 50 + 25 + 12 + 6 + 3 + 1 = 97
Now we will find
Highest power of 2 in 10000! = [10000/21]  + [ 10000/22] + [10000/23] + [10000/24] + [10000/25] + ..... + [10000/213]
                                          = 5000+2500+1250+625+312+156+78+39+19+9+4+2+!
                                          = 9995                                                   
Hence Highest power of 297in 10000! = [9995/97] = 103

Hence the maximum value of k would be 103 and not 104.
I hope this post will help some.

Thanks
SatyamGadha


Re: Divisibility ...to Deepak Kansal
by praveen kumar - Wednesday, 2 July 2008, 04:29 PM
 

A slight mistake in your arguement .

similarly for 34^333 = (36-2)^333

all the terms will be divisible by 9 except     (-2)^333 = -(2)^333, since 333 is odd.

Net remainder will be 2.

The number is divisible by 11.

Re: Divisibility
by sanjeev ganesh - Wednesday, 13 August 2008, 04:47 AM
 

Thanks Satyam,

 

I am not clear with below step of urs:

Now we will find
Highest power of 2 in 10000! = [10000/21]  + [ 10000/22] + [10000/23] + [10000/24] + [10000/25] + ..... + [10000/213]
                                          = 5000+2500+1250+625+312+156+78+39+19+9+4+2+!
                                          = 9995                                                   
Hence Highest power of 297in 10000! = [9995/97] = 103

Can you please explain me, why have you performed this step. Is it possible to apply for (100!)=(10!)^k+p ?

 

~Sanjeev


Re: ques !!!!
by Sibsankar Dasmahapatra - Friday, 12 September 2008, 04:37 PM
 

Hi Dalal / TG

                            I think this problem u can do by the following process......

N= 1*2+2*3+.....+99*100 ; So N= n*(n+1)=n^2+n..

So sum of N =  Sum of (n^2) + Sum of (n)

                     = n*(n+1)*(2n+1)/6  +  n*(n+1)/2

here n=100...so sum of N =(100*101*201)/6 +  (100*101)/2...that is divisible by 101...

i think i am correct..

Thanks

Sibu

Re: Divisibility
by Sibsankar Dasmahapatra - Friday, 12 September 2008, 06:16 PM
 

Hi Sanjeev

                        Let me explain your first query Highest power of 297in 10000!.....i will give u a simple example...suppose u have to find out highest power of 2 in 4!..4! means 1.2.3.4 = 24 = 3.8 = 3.2^3...so highest power of 2 is 3 ; now apply the theory to find out the highest power of 2 in 4!...4/2+4/2^2 = 3 ...this the shortcut process to find the max power .....

now if u want to find out the max power of 4 in 4! then ; 4 means 2^2 ....so u have to find out the max power of 2 , then you divide the the max power of 2 by 2 (as 4=2^2)...if u have to find the max power of 8 in 4! ; then 8 means 2^3...so find the max power of 2 then divide by 3 .....do the same thing in above question...

..

Now your second query Is it possible to apply for (100!)=(10!)^k+p ?...yes it is applicable....in this case find the max power of 7 (as it is max prime number in 10 !) in 100! and max power of 2^8 (as it is the max power in 10! )  in 100!

 

 

Thanks

Sibu

Re: Divisibility
by abdul rasheed - Tuesday, 3 February 2009, 04:22 PM
 

43 ^ 444 = (45-2)^444

all terms will be divisible by 9 except 2^444.

2^444 = (9-1)^148

remainder is 1

similarly for 34^333 = (36-2)^333

all the terms will be divisible by 9 except 2^333

ABDUR: It should be "all the terms will be divisible by 9 except (- 2^333 )"

2^333 = (9-1)^111

ABDUR: It should be " -2^333 = - (9-1)^111

remainder is -1 or 8

ABDUR: It should be "Remainder 1 ".

if we add both the remainders thus 8+1 = 9

hence divisible by 9

Re: Divisibility
by shivam mehra - Monday, 9 February 2009, 12:38 AM
  to cut long story short

divisibility test  N=abcde

7         abcd-2e  and keep on solving till ge a 2/1 digit number
13       abcd+4e  and keep on solving till ge a 2/1 digit number
17       abcd-5e  and keep on solving till ge a 2/1 digit number
19       abcd+2e  and keep on solving till ge a 2/1 digit number

eg 2401
240-2(1)=238
23-2(8)=7

2197
219+4(7)=219+28=247
24+4(7)=52
5+4(2)=13

4913
491-5(3)=476
47-5(6)=17

6859
685+2(9)=703
70+2(3)=76
7+2(6)=19


shivam
Re: Divisibility
by Deepika Khandelwal - Tuesday, 17 February 2009, 03:28 PM
  hi deepak
can you explain how 8 reminder is coming for 34^333
in question 43^444+34^333

Thanks
Re: Divisibility
by vikas sharma - Wednesday, 25 February 2009, 04:56 PM
 

Hi harshal as i think

like take n= 123456789 as one digit so nnnn-----20 times

any same digit no is divisible by 7/11/13 if no of digits multiple of 6

here last 2 digits are spare(nn)

so check only divisiblity of 123456789123456789 and its divisible by 11

so remainder 0

m i right TG sir

Re: Divisibility
by vikas sharma - Wednesday, 25 February 2009, 05:20 PM
 

hi Satyam can u eloborate it more, i didnt get it

i know this concept -the power m in n! is calculated by [n/a]+[n/a2] and so on where a is highesst prime of m.

 

Re: Divisibility
by bhavika bali - Tuesday, 10 March 2009, 01:35 PM
 

doubt!!can u plzzz help me..

q)a number when divided by 5 gives a number which is 8 more than remainder obtained on dividing the same number by 34 .such least possible number is??

a)175     b)75  c)680   d)does not exist

q)when a number "N" is divided by a proper divisor  "d" then it leaves a remainder of 14 and if thrice of that number ie "3N",is divided by same divisor d,the reaminder comes out to be 8.again if 4 times of the same number ie "4N", is divided by d remainder will be??

A)35     B)22     C)5  D)cant be determined

Re: Divisibility
by dimpu n - Wednesday, 11 March 2009, 12:25 PM
  when a number "N" is divided by a proper divisor  "d" then it leaves a remainder of 14 and if thrice of that number ie "3N",is divided by same divisor d,the reaminder comes out to be 8.again if 4 times of the same number ie "4N", is divided by d remainder will be??

A)35     B)22     C)5  D)cant be determined

N divided by d  remainder is 14

N =  dx+14

3N=3dx+42

3N divided by d remainder is 8

which means 42 divided by d gives aremainder of 8

ie d should be a factor of 34 & d >8 &d >14

d can be 17 or 34

d is 17 remainder when N is divided by d is 5 and 22 if d is 34

so the answer is cannot be determined

Please correct me if i am wrong  

Re: Divisibility
by Varun Agrawal - Thursday, 21 May 2009, 05:35 PM
  Ques 1.  75 is the answer. put n see approach smile 
Re: Divisibility
by Deepika Khandelwal - Thursday, 28 May 2009, 10:54 AM
  Hello TG/colleagues,
could you please explain how to find out that this no. is
perfect sq.

ex. AB36 is perfect sq. or not
How to solve thsi type of questions.

Thanks

Re: Divisibility
by dibya ranjan pal - Wednesday, 3 June 2009, 06:39 PM
 

I think here the principle of digit sum is required....

according to it a square's digit sum has to be 1,4,7 or 9....

 image

A number will NOT be a perfect square if its digit-sum is NOT 1, 4, 7, or 9, but it may or may not be a perfect square if its digit-sum is 1, 4, 7, or 9.

Re: Divisibility
by Ankit Megotia - Monday, 20 July 2009, 11:15 AM
  hi deepika,
for a number to be a perfect square the last digits should be one of 1,4,5,6 and 9. also, a perfect square has its ten's digit even except when the unit digit is 6. therefore AB36 can or cannot be a perfect square.
Re: Divisibility
by saumya agarwal - Wednesday, 22 July 2009, 12:19 AM
    10000!=(100!)^k A-P?? im not able to get the solution given by tg ...the one which says dat we need to compare the powers.....can anyone explain  me this method once again...with another example..
Re: Divisibility
by shailesh jain - Thursday, 23 July 2009, 02:29 PM
 

Hi,

IF x lies between 1 and 1000 both inclusive, then how many values are there between them so that the equation 4x^6 + x^3 + 5 is divisible by 7.

Please let me know if any one has solved the problem.

Thanks

Re: Divisibility
by Netra Mehta - Friday, 4 September 2009, 01:40 PM
  Hi Bhavika!

The ans 4 ques 1 is 75..check it thru the options n u ll get..

n for question 2...I hv done lk dat

Let N=d*q1 +14 -- (1)
3N=d*q2 +8 -- (2)
4N=d*q3 + r -- (3)

whr q1,q2,q3 are all integers

By (3) - (2) v get
N=d(q3-q2)+(r-8) -- (4)

now by equating (4) & (1) v can write..

r-8 = 14
& therefore r= 22

This is how I got the answer...
Tell me if m wrong...

Re: Divisibility
by rupasree raj - Saturday, 5 September 2009, 08:58 AM
 

q)a number when divided by 5 gives a number which is 8 more than remainder obtained on dividing the same number by 34 .such least possible number is??

a)175     b)75  c)680   d)does not exist




i don't know if its correct, but i got de ans as 75 by dis approach.
ans:    let 'n' be the no: dats being divided by 5. let de quotient be 'x' and de remainder be 'y'. now its said dat 'n' is again divided by 34. let de quotient be 'q' and remainder be 'r'.
                         also, x=r+8
                        5x+y=n..............(1)
               and    34q+r=n.............(2)
equate both de eqns.
     ie; 5x+y=34q+r.....................(3)
to get de least value of 'n' , 'y' shud b zero. also 'q' shud get de least value.
 lets put q=1.
             we'll get eqn(3) as: 5x=34+r
                               now we know x=r+8
                            5r+40 = 34+r
wen v solve dis v'll get a negative ans for 'r'. so v can now change de value of 'q' as 2(de next least no). nw eqn(3) becomes:
                             5(r+8)=34*2+r
                             5r+40=68+r
on solvin v get: r=7.
so: x=15
n=15*5=75


pls correct if its wrong... thank u.....
Re: Divisibility
by neha saxena - Saturday, 5 September 2009, 09:53 AM
  hi Biswajit

The question here is to find the smallest natural number...33 is a probable answer in case the quest says that find a natural number such that its factorial is divisible by 1089,  your solution will also satisfy.
But focus on the word smile

the concept i used is 1089 = 3 ^2 * 11 ^2
thus if we consider 22!
we can have 11 and 22 two numbers divisible by 11^2.
thus the answer should be 22.
I hope you wud have got it correct
Re: Divisibility
by prajay desai - Saturday, 5 September 2009, 11:38 AM
  hi TG SIR......
PLZ CAN U TEL ME D METHOD??????
12341234..........upto 400 digits.......find the remaindr when it is divided by 101...
i cant undrstand the chineese method so plz can u explain it in a easy way?????
Re: Divisibility
by Sugato Ray - Saturday, 5 September 2009, 08:00 PM
  Hi Prajay,

12341234....upto 400 times (i.e 1234 ... 100 times)
=> 1234x100198+1234x100196+1234x100194+....+1234x1004+1234x1002+1234x1000 = N (say)

Now Remainder(1234/101) => 22.
=> Remainder(N/101) = Remainder[(22x100198+22x100196+22x100194+....+22x1004+22x1002+22x1000)/101]
= Sum over x on {Rem[22x1002x+/(100+1)]}..... x=198/2,196/2,...,0  (100 terms)

Rem[22x1002x+/(100+1)] = Rem[22*y2x+/(y-(-1))] ... y = 100
= { 22*(-1)2x} ... using remainder theorem

Therefore, Remainder(N/101) = Rem[(Sum over x on { 22*1})/101] ... for 100 terms of x
= Rem(22x100/101) = 22*(-1) = -22 = 101-22 = 79

Please correct if I am wrong!... smile
Re: Divisibility
by manoj kumar - Saturday, 5 September 2009, 09:21 PM
  Hello TG sir/ guys..

one easier method for divisibility test of 7 or 11

For 7:
say the number is 343..(we know it is divisible by 7.just an example)
Split the number into two parts one 34!3 i.e one part is 3 nd other part is remaining number.
take the last digit(3 in this case) .multiply it with 2 (3 *2 = 6) subtract 6 from the remaining number i.e from 34 i.e 34 - 6 = 28..28 is divisible,so is 343.
one bigger number 117649..divide number into two parts.11764!9 take the last digit , multiply by 2 nd subtract from remaining..repeat this until u r satisfied that the resultant is divisible by 7..

11764 - (9*2) = 11746 (duno if this is divisible or not) so repeat..
now number becomes 11746..split 1174!6 ..1174-(6*2) = 1162...repeat
1162..split 116!2.. 116-(2*2) = 112 ..this i knw is divisible by 7..hence the original number 117649 is also divisible by 7...

there r similar methods for all prime numbers ...nd that too wid a general formula..will post them soon...
Re: Divisibility
by shobhit agarwal - Sunday, 11 April 2010, 09:40 AM
  Hello TG sir/guys

could anyone please explain how to find out whether a number is perfect sq. or not?
Re: Divisibility
by vivek bhatt - Saturday, 7 January 2012, 09:52 PM
  i didnt get this can u please explain the sum in mch simpler way please
Re: Divisibility
by gaurav prakash - Tuesday, 24 January 2012, 11:15 PM
  in the above solution

43 ^ 444 when divided by 9 gives

remainder 1

BUT
i think

34^333 = (36-2)^333 will give remainder 1 instead of 8

all the terms will be divisible by 9 except (-2)^333

= (-8)^111

when -8 is divided by 9 it gives remainder 1
total remainder will be 1+1=2

hence not divisible by 9

please mark me if i m wrong :-)
Re: Divisibility
by Ani Rai - Wednesday, 25 January 2012, 02:22 PM
  find the number of integer solution to the equation 1/x + 1/y = 1/48
Re: Divisibility
by TG Team - Friday, 27 January 2012, 12:15 PM
 

Hi Ani smile

Let x = 48 + a

and y = 48 + b

so 1/x + 1/y = 1/48

=> 1/(48 + a) + 1/(48 + b) = 1/48

=> 48(48 + a) + 48(48 + b) = (48 + a)(48 + b) = 48² + 48a + 48b + ab

=> ab = 48² = 28

Now we just need to find the number of ways in which RHS can be written as product of two integers.

See, number of (positive) factors of RHS = (8 + 1)(2 + 1) = 27.....so number of (ordered) pairs of positive factors possible = 27.

And including the negative values for a, b we get total 2 × 27 = 54 pairs. smile

Kamal Lohia 

Re: Divisibility
by Abhishek Sharma - Wednesday, 15 February 2012, 08:35 PM
 

Hi Ankit,

As u told the rule for checking the no is a perfect sqaure  or not Then what should be sol for following question: A 10-digit number N has among its digits one 1, two 2s, three 3s, and four 4s. Is N be a perfect square?Plz explain

Re: Divisibility
by Abhishek Sharma - Wednesday, 15 February 2012, 08:41 PM
 

Hi All,

Plz explain how to solve this question(From your quiz)

All possible pairs are formed from the divisors of 21600. How many such pairs have HCF of 45?

Re: Divisibility
by TG Team - Thursday, 16 February 2012, 04:16 PM
 

Hi Abhishek smile

Refer to digit-sum property of perfect squares. For article - click here.

Kamal Lohia

Re: Divisibility
by tushar varshney - Friday, 2 November 2012, 12:21 PM
  hi Biswajit sen

yes u r wrong
because in this text we have asked for the least value
the factors of 1089 is
3^2*11^2
so 3^2 can be come in every factorial greater than or equal to 6
now we have to concentrate on 11
we have need 11 at most 2 times
which will come in 22
which is least value
Re: Divisibility
by Amit Lawange - Tuesday, 20 November 2012, 09:29 PM
  Hi

I didnt understand how 2^444 became (9-1)^148 nd similarly (9-1) ^111 could you please elaborate it for me....Thnx in advance
Re: Divisibility
by tgdel640 tg - Wednesday, 21 November 2012, 04:32 PM
  2^444 = (2^3)^148
= 8^148
= (9-1)^148

Hope this helped.

Regards
Ankit Gaur
(TathaGat)
Re: Divisibility
by Yashodhan Bhatt - Tuesday, 6 May 2014, 11:59 AM
  Can someone please explain what is meant by 888…888A999…999 ?
I mean, what does this symbol … stand for in this number ?
Re: Divisibility
by TG Team - Tuesday, 6 May 2014, 06:14 PM
  Hi Yashodhan smile

It is simply font problem of the system....actually the number mentioned here is: 888....888A999....999 where both 8's and 9's are written 50 times.

I hope it is clear. smile

Kamal Lohia
Re: Divisibility
by Yashodhan Bhatt - Thursday, 8 May 2014, 10:46 AM
  Thank you sir for the timely reply smile

Re: Divisibility
by Yashodhan Bhatt - Thursday, 8 May 2014, 10:48 AM
  Sir,
Would the team plan & post the preparation study plan for CAT'14 as TG sir did earlier in the following link ?

http://totalgadha.com/mod/forum/discuss.php?d=2619&mode=-1

I have even requested TG sir the same question as now for CAT'14 we do have to consider the past 5 years (2009 onwards) question papers.

Awaiting your reply.

Regards
Re: Divisibility
by TG Team - Tuesday, 13 May 2014, 01:27 PM
  Hi Yashodhan smile

Study plan remains same as types of questions and topics of the question being asked in previous 5 years of CAT has not changed much. Little bit level of difficulty has been varied. Further, previous 5 year CAT were held on computer, so question papers were not provided after the exams. So it is not possible to do their analysis in same manner as done in the link posted by you.

But conclusive suggestion for you is to follow the similar plan as advised in earlier link, that'll give you a good boost up.

All the best for your prep journey. smile

Kamal Lohia