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Last Digit (Posted on 2004-01-23) Difficulty: 3 of 5
Find the last digit of summation of the series:
(1)^99 + (2)^99 + (3)^99 + (4)^99 + + (98)^99 + (99)^99

See The Solution Submitted by Ravi Raja    
Rating: 2.4000 (5 votes)

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Solution Last two digits | Comment 15 of 19 |
By the remainder theorem (also known as the factor theorem), non-zero (a - b) divides (a^n - b^n). For odd n, we therefore have (a - (-b)) divides (a^n - (-b)^n), or (a + b) divides (a^n + b^n).

For odd n, we can also explicitly factorize a^n + b^n, as follows. Writing n = 2m+1, we have:
a^(2m+1) + b^(2m+1) = (a + b)(a^2m - a^(2m-1)b + a^(2m-2)b^2 - ... - ab^(2m-1) + b^2m).

Rewriting the series as (1^99 + 99^99) + (2^99 + 98^99) + ... + (49^99 + 51^99) + (50^99), 100 divides each of the first 49 bracketed terms by the factor theorem. Since 100 divides 50^2, it also divides 50^99.

Therefore 100 divides the whole series, and the last two digits are 00.

Edited on January 31, 2004, 7:53 am
Edited on January 31, 2004, 8:06 am
  Posted by Nick Hobson on 2004-01-30 15:22:31
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