What is the remainder when the number is divided by 99?
a. 18  
b. 33  Super!  
c. 36  
d. 27 
quiz problem./....Number system..  
6
Marks: 1.33/2 What is the remainder when the number is divided by 99? Choose one answer.

Re: quiz problem./....Number system..another question  
Check out the following pattern: 33*66=2178 [3rd digit from right is 1] Hence, answer is 1 
Re: quiz problem./....Number system..another question  
Thanx buddies.... solve the second ques also... 
Re: quiz problem./....Number system..another question  
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  
All possible pairs are formed from the divisors of 21600. How many such pairs have HCF of 45? Choose one answer.
** 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  
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. Srini 
Re: quiz problem./....Number system..another question  
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  
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..  
123 {1001001001....1001} (298 digits) = 123 {10^{297} + 10^{294} + 10^{291} + .... + 10^{3} + 1} (100 terms) The remainder of 10^{odd} is 10 with 99 and the remainder of 10^{even} 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 Ã— 50 + 1 Ã— 50} = 24{550} = 24 Ã— 55 = 1320. Remainder of 1320 with 99 = 33. 
Re: quiz problem./....Number system..another question  
The pair will be (45a, 45b) where a and b will be coprime
to each other. Now 21600 = 2^{5} Ã— 3^{3} Ã— 5^{2}. To find a and b, we first take the factor
of 45 from 21600, which leaves 2^{5} Ã— 3 Ã— 5. Now
we need to find the number of coprime pairs (a, b) that we can make out of 2^{5} Ã— 3 Ã— 5. Letâ€™s
write down the powers of the prime factors in order to find the coprime
factors:
(2, 2^{2}, 2^{3}, 2^{4}, 2^{5}), 3, 5 Therefore, the number of coprime pairs is found by
various combinations of these prime factors: (Prime factor, Prime factor) â€“ (2,
3), (2^{2}, 3), (2^{3}, 3)â€¦ (2^{4}, 5), (2^{5},
5), (3, 5)  11 in number Therefore, total number of coprime pairs (a, b) = 49. Total Gadha 
Re: quiz problem./....Number system..another question  
Hi srini vasan, we have only 300 digits, and not 300 sets. 
Re: quiz problem./....Number system..another question  
yes... I missed that... Srini 
Re: quiz problem./....Number system..another question  
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 7224+1=49 so 49 factors is ans 
Re: quiz problem./....Number system..another question  
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... regards, fundoo 
Re: quiz problem./....Number system..another question  
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  
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  
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 