Find the last digit of summation of the series:
(1)^99 + (2)^99 + (3)^99 + (4)^99 + ……… + (98)^99 + (99)^99

zero
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Only the one's digit is relevant here, so 1 is congruent to 11 is congruent to 91, etc... and the zeroes (10, 20, 30, etc...) all end in zero... so we can ignore them.

So, for each of the nine 'congruence sets' such as 1, 11, 21, 31, ... 91 through 9, 19, 29, etc... each member in the set, raised to 99th power, end in the same digit.

And all nine of the 'congruence sets' have 10 elements.... since we're adding them all up 10x(ANY digit) ends in zero.

So, the total summation must end in zero.

Posted by SilverKnight on 2004-01-23 09:47:03