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quiz problem./....Number system..
by total senti - Tuesday, 17 July 2007, 12:35 PM
 
6
Marks: 1.33/2

What is the remainder when the number imageis divided by 99?

Choose one answer.
a. 18
b. 33
c. 36
d. 27
Re: quiz problem./....Number system..
by Abhishek Jain - Tuesday, 17 July 2007, 01:49 PM
 

in this problem u need to first divide the divisor into two halves i.e., 9 nd 11 after that u need to apply the divisibility rule of both 9 nd 11 on the number.  the operation is as follow-

since every number is divisible by 9 whose sum of digits taken three at a time is divisible by 9. for eg 135540 is divisible by 9 coz 135+540 is divisible by nine. v apply the same rule here.

nd therefore v add 123 + 123 + ........ 100 times which is nothin but 123*100 which when divided by 9 leaves the remainder 6.

hence the number is of the form 9m + 6.   (a)


another divisibility rule is that of 11. wherein v first subtract nd then add d digits taken three at a time. for eg 123456789 when divided by 11 will give u the same remainder as 789-456+123 when divided by 11.

v apply this rule in d above question nd get

(123-123)+(123-123)............... nd so on. since there are even no of such pairs so their sum turns out to b 0 nd hence the no is divisible by 11.

therefore the number is of the form 11n.          (b)

by computing the two results a nd b

v get the equation 9m +6 = 11n

nd by using special equation v figure out that the first number satisfying the equation is 33. nd therefore 33 is the remainder

 

Re: quiz problem./....Number system..another question
by total senti - Tuesday, 17 July 2007, 03:17 PM
 

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

Choose one answer.
a. 49
b. 34
c. 8
d. 276

*****

Let A = 333…333 (51 digits) and B = 666…666 (51 digits). Then the 52nd digit (counting from the right) in the product A Ã— B is
Choose one answer.
a. 7
b. 2
c. 9
d. 1
Re: quiz problem./....Number system..another question
by Small Wonder - Tuesday, 17 July 2007, 03:34 PM
 

Check out the following pattern:

33*66=2178 [3rd digit from right is 1]
333*666=221778 [4th digit from right is 1]
3333*6666=22217778 [5th digit from right is 1]
33333*66666=2222177778 [6th digit from right is 1]
Same pattern continues.

Hence, answer is 1

Re: quiz problem./....Number system..another question
by total senti - Tuesday, 17 July 2007, 03:57 PM
  Thanx buddies.... solve the second ques also...
Re: quiz problem./....Number system..another question
by brendan decruz - Tuesday, 17 July 2007, 04:13 PM
  hi,
q1. is it option d? actually i am getting answer greater than first three options.
so am asking...... but if its so, i will try and provide the solution towards getting it....
q2. option d.1  (i.e. counting from the right).
Re: quiz problem./....Number system..another question
by total senti - Tuesday, 17 July 2007, 04:38 PM
 

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

Choose one answer.
a. 49
b. 34
c. 8

**

my approach: factors can be:  5^2  * 3^3  * 2^5,

so total no of possible factors are 3*4*6 = 72

now we nedd to pair them up.... so 36, now answer shd be between 8/34.... 34 is likely..... but nyways this cant be a gud approach.... help me out guys......  TG Boss whr r u......??

 

Re: quiz problem./....Number system..another question
by srini vasan - Tuesday, 17 July 2007, 04:44 PM
  One doubt...

123123123123123123 - 6 sets. These are both divisible by 11  and 9 hence it is divisible by 99.

For 300/6 = 50 such sets. so the remainder is 0 any ways.
I solved this some time back and got the answer as 33. Now got this doubt.sad

-Srini
Re: quiz problem./....Number system..another question
by Ranvijay Singh - Tuesday, 17 July 2007, 09:14 PM
  Answer to Q2: 49
Approach: 21600 = 2^5 * 3^3 * 5^2
For HCF of 45, we need to have 3^2 * 5 in both the terms of the selected pair. Now, other seven factors (2^5, 3, 5) can be multiplied to each of the terms in several ways giving the HCF of 45. Total no. of all such pairs would be (7 * 1 + 6 * 2 + 5 * 3 + 4 * 4 - 1) = 49.

Note: -1 becoz there would be one repetition.


Regards,
~Vijay
Re: quiz problem./....Number system..another question
by Arun Reddy - Tuesday, 17 July 2007, 11:48 PM
 

Hi Vijay,

Could you please elaborate your approach...I could not understand below step

(7 * 1 + 6 * 2 + 5 * 3 + 4 * 4 - 1)

Thanks,

Arun.

Re: quiz problem./....Number system..
by Total Gadha - Wednesday, 18 July 2007, 01:30 AM
  123 {1001001001....1001} (298 digits)
= 123 {10297 + 10294 + 10291 + .... + 103 + 1} (100 terms)
The remainder of 10odd is 10 with 99 and the remainder of 10even is 1 with 99. There are 50 terms with odd powers and 50 terms with even powers. Also, 123 gives remainder of 24 with 99.
Therefore, remainder = 24{10 × 50 + 1 × 50} = 24{550} = 24 × 55 = 1320. Remainder of 1320 with 99 = 33.
Re: quiz problem./....Number system..another question
by Total Gadha - Wednesday, 18 July 2007, 01:32 AM
 

The pair will be (45a, 45b) where a and b will be co-prime to each other. Now 21600 = 25 × 33 × 52. To find a and b, we first take the factor of 45 from 21600, which leaves 25 × 3 × 5. Now we need to find the number of co-prime pairs (a, b) that we can make out of 25 × 3 × 5. Let’s write down the powers of the prime factors in order to find the co-prime factors:

                                                (2, 22, 23, 24, 25), 3, 5

Therefore, the number of co-prime pairs is found by various combinations of these prime factors:

(Prime factor, Prime factor) – (2, 3), (22, 3), (23, 3)… (24, 5), (25, 5), (3, 5) ----- 11 in number
(Two prime factors, prime factor) – (2 Ã— 3, 5), (22 × 3, 5),… (24 × 5, 3), (25 × 5, 3), (3 × 5, 2), … (3 × 5, 25)  --- 15 in number
(1, prime factor) – (1, 2), (1, 22) …(1, 25), (1, 3), (1, 5) --- 7 in number
(1, two prime factors) – (1, 2 × 3), (1, 22 × 3), … (1, 24 × 5), (1, 25 × 5), (1, 3 × 5) --- 11 in number
(1, three prime factors)- (1, 2 × 3 × 5), (1, 22 × 3 × 5), … (1, 25 × 3 × 5) --- 5 in number.

Therefore, total number of co-prime pairs (a, b) = 49.


Total Gadha
Re: quiz problem./....Number system..another question
by Sri KLR - Wednesday, 18 July 2007, 09:31 AM
 

Hi srini vasan,

we have only 300 digits, and not 300 sets.

Re: quiz problem./....Number system..another question
by srini vasan - Wednesday, 18 July 2007, 03:07 PM
  yes... I missed that...
-Srini
Re: quiz problem./....Number system..another question
by shiva chepuri - Monday, 30 July 2007, 10:34 PM
  hi the solutions of first problem  can be
2^5*3^3*5^2=so there are 72 factors(6*4*3)
45=3^2*5
so when 21600 is divided by 45 the no is 2^5*3*5
this no has 24 factors
so no of pairs with hcf as 45 is 72-24+1=49
so 49 factors is ans
Re: quiz problem./....Number system..another question
by fundoo bond - Tuesday, 31 July 2007, 12:27 PM
 

hi ranvijay,

   though i understood the expl given by TG,i want to knw wht "(7 * 1 + 6 * 2 + 5 * 3 + 4 * 4 - 1) = 49."  this step was all about? can u plz generalize it for me... smile

regards,

fundoo

Re: quiz problem./....Number system..another question
by Sri KLR - Tuesday, 31 July 2007, 12:34 PM
 

Hi TG,

for this kind of problem you have given a formula....

It was something like (p+1)(q+1)(r+1)-pqr..

I read some time back....but now I couldn't find...plz help.

Re: quiz problem./....Number system..another question
by arsenal . - Sunday, 7 October 2007, 04:36 AM
 

Its something like this i suppose

[ (p+1)(q+1)(r+1) - 1 ] + pq + qr + pr + 3pqr

can anyone confirm

Re: quiz problem./....Number system..another question
by shelly verma - Monday, 16 November 2015, 07:16 PM
  hi ,

In order to get 45 in both the terms of a pair , 21600 should contain two 45's and here 21600 contain only one 45 ..
 (45a , 45b ) =21600 where a,b are co primes .

I cannot proceed from here . Pls. help