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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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Some Thoughts re(2): Last two digits | Comment 17 of 19 |
(In reply to re: Last two digits by Richard)

"But if I take a=3, b=1, and n=2, a^n+b^n=10, while a+b=4 and 4 does not divide 10. It is true that a-b divides a^n-b^n."

Thanks Richard, you're right, I was a little hasty there! I meant to say that (a + b) divides (a^n + b^n), for *odd* n. This does follow from the remainder theorem, and I've edited my post accordingly.
Edited on January 31, 2004, 7:59 am
  Posted by Nick Hobson on 2004-01-31 07:57:31

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