## Re: [2IIM CAT Prep] some doubts

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• Hello Sri, 1. either u can add 91...99 using AP n then divide by 6. or u can individually divide each no by 6 n then add the remainders, n repeat the proc
Message 1 of 2 , Dec 28, 2007
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Hello Sri,
1. either u can add 91...99 using AP n then divide by 6.
or u can individually divide each no by 6 n then add the remainders, n repeat the proc untill u get the remainder < 6.
2. i think, no option but to calculate using some shortcuts. like, in each century(< 1000), 2 appears 22 times n so on.
3.here, actually, we have to find value of n when 1^n and 2^n give the same remainder. (351 and 352 when divided by 7 give remainder 1 n 2 respectively.) Ultimately we have to find that power of 2 which will give remainder 1. (because, for any value of n, 1^n will always give rem as 1.

On Dec 19, 2007 12:31 AM, sri rangan <ranganece@...> wrote:

 1  What is the reminder when 91 + 92 + 93 + ...... + 99 is divided by 6?    2   How many times will the digit '0' appear between 1 and 10,000?     3  For what value of 'n' will the remainder of 351^n and 352^n be the same when     divided by 7?                  CAN ANYONE EXPLAIN ME THESE CLEARLY......