I need proper explaination. OA is B

n/7 = q+r n = 7q+r (q->quotient r->remainder) Whats is r? (could be 0,1,2,3,4,5,6) Stmt I n/21 is odd integer(1,3,5,7....) Consider remainder =1 ,3 case n/21 = q+1 or n/21 = q + 3 and so on.... n=21q+1 or n= 21q+3 Take q=0 n=1 or n=3 Divide by 7 gives 2 remainder values INSUFF Stmt II n/28 = q+3 n=28q+3 q=0 n=3 3/7 remainder 3 q=1 n = 31 31/7 remainder 3 q=2 n=59 59/7 remainder 3 q=3 n = 87 87/7 remainder 3 It follows a pattern where the remainder will always be 3 ONE DEFINITE ANSWER SUFF B) Regards, Cramya

i had an idea. if a number say a when divided by b leaves a remainder r, then a when divided by any factor of b will leave the same remainder r. can we apply this as a rule? in the above: n/28 left a remainder 3. so 7 is a factor of 28. so n/7 would also leave 3 as remainder, as would n/2, n/4, n/14 etc

I had an idea. if a number say a when divided by b leaves a remainder r, then a when divided by any factor of b will leave the same remainder r. can we apply this as a rule? U read my mind. I was about to post it and leave it open for other comments. The only thing I would add is factors that are greater than the remainder in question( i.e 3 in this problem) . Eg: 28 has 6 factors 1,2,4,7,14,28 If u look at 4,7,14,28 all leave a remainder 3 for the same n 2 leaves a remainder 1 1 leaves a remainder 0 Regards, Cramya

thanks cramya. got it. No problem; good luck with ur prep Scoobydooby! Nice thinking on your part also!

Thnks guys.. appreciate ur help..

Nice question: Given N = 7 * K + r -----------{1} r CANNOT be greater than 7 option: 2 N = 28 m + 3 ------------{2} 28m + 3 = 7*K + r r = 7* t + 3 r = { 3, 10, 17 , ...) but r should be less than 7 so r is 3 Option 1 r = 7*t + ODD no. which can be 5 or 3

from the stem we know we have to find out the remainder when divided by 7 ! A) its an odd number ! number can be 23,25,27 all divided by 21 gives odd number B)n=31 31/28=3 n=59 59/28=3 31/7 remainder is 3 59/3 again fetches u 3 B it is