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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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re(3): solution | Comment 9 of 19 |
(In reply to re(2): solution by Ady TZIDON)

The way to edit one's comments is to select that comment, and below the comment's text will be a place where it says "This is your comment". Near that is a link that says "Edit comment". But this feature is generally looked upon as to be used only to correct typos, so a follow-up note is good as a general practice anyway (as I did in correcting my leaving out of the digit 3 in my analysis).

Regarding your correction to
"Remark(NOT NEEDED TO SOLVE THE PUZZLEW): the last digit of Nth power of any number is equal to the last digit of Kth power of said number iff(=if and only if) M mod 5= N mod 5 i.e. consider the 4th power instead of the 99th."

First I assume you mean K rather than M as M has not been introduced. And your correction indicates mod 4 rather than mod 5. So it would read

the last digit of Nth power of any number is equal to the last digit of Kth power of said number iff(=if and only if) K mod 4 = N mod 4 i.e. consider the 3rd power instead of the 99th.

Actually the word should be "if" rather than "iff", as K = N mod 4 (which is another way of saying K mod 4 = N mod 4) is sufficient for the last digits to be the same, but not always necessary. For example, 4^3 ends in 4 as does 4^5 even though 3 and 5 have different congruences mod 4. The "iff" holds in both directions only for numbers ending in 2, 3, 7 and 8; the other digits have cycles of length 1 or 2.

  Posted by Charlie on 2004-01-23 10:38:36

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