anand bhatnagar 0 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 post Share on other sites

2 DejMar 9 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 post Share on other sites

0 jasen 4 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 post Share on other sites

0 mmiguel 1 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 post Share on other sites

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## anand bhatnagar 0

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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