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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)

Comments: ( Back to comment list | You must be logged in to post comments.)
re: Last two digits | Comment 16 of 19 |
(In reply to Last two digits by Nick Hobson)

"By the remainder theorem (also known as the factor theorem), non-zero (a + b) divides (a^n + b^n)."

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.
  Posted by Richard on 2004-01-30 22:09:15

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