anand bhatnagar Posted August 5, 2016 Report Share Posted August 5, 2016 1^1+2^2+3^3+4^4........ +99^99 what are the last two digits of the above expression ? and why...? Quote Link to comment Share on other sites More sharing options...

2 DejMar Posted August 5, 2016 Report Share Posted August 5, 2016 (edited) Spoiler 20. The last two digits each n^{n} number follows a specific pattern. There is likely a more simple way of addressing this pattern... If the final (ones) digit of n were 0, 1 or 9, then the final digit of n^{n} is respectively 0, 1 or 9. If the final digit of n were 2 or 8, then the final digit of n^{n} is respectively 4 and 6 where the tens-digit is even,or 6 and 4 if.odd. If the final digit of n were 3 or 7, then the final digit of n^{n} is respectively 7 and 3 where the tens-digit is even,or 3 and 7 if.odd. In each of the above cases, the tens-digit is represented respectively by each different numeral, i.e, {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} for each n. The sum of these ten digits will be 45, and with the tens digit of the sum being 0 (5 × 6 = 30) as the left-most digit of the six pairs is a hundreds digit. If the final digit of n were 4 or 6, then the final digit of n^{n} is respectively 6. The tens-digit, in both follows the sequence {5, 1, 7, 3, 9}. The sum of the four sets of the quintuplets of tens-digits is (4 × (5+1+7+3+9) = 200). And the sum of the twenty ones-digits ending in 6 is 120. The total sum for the final digits of 4 and 6, then is 320; that is, the ones-digit is 0 and the tens-digit is 2. If the final digit of n were 5, then the final digit of n^{n} is respectively 5. The tens-digit of n^{n} alternates between 2 for the even tens-digit of n and 7 for the odd tens-digit of n. Given the five pairs, the sum of these ten is 500, that is, the ones-digit and tens-digit are both 0. Thus, by summation, the ones-digit in the final solution must be 0. and the tens-digit must be a 2. Edited August 5, 2016 by DejMar correction of typo Quote Link to comment Share on other sites More sharing options...

0 jasen Posted August 6, 2016 Report Share Posted August 6, 2016 (edited) Spoiler Edited August 6, 2016 by jasen suddenly realize wrong answer Quote Link to comment Share on other sites More sharing options...

0 mmiguel Posted September 6, 2016 Report Share Posted September 6, 2016 Spoiler 20 because python def nnmod100(n): nm = n % 100; p = 1; for i in xrange(n): p = (p*nm) % 100; return p; s = sum(nnmod100(n) for n in range(1,100)); print s % 100; Quote Link to comment Share on other sites More sharing options...

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## anand bhatnagar

1^1+2^2+3^3+4^4........ +99^99

what are the last two digits of the above expression ? and why...?

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