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.**so n=3 is the answer.**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......

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Thanks and Regards,

Nachiket Joshi

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