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

 Submitted by Ravi Raja Rating: 2.4000 (5 votes) Solution: (Hide) We group the sum as follow: [(1)^99 + (11)^99 + ... + (91)^99] + [(2)^99 + (22)^99 + ... (92)^99] + ...... + [(9)^99 + (19)^99 + ... + (99)^99] + [(10)^99 + (20)^99 + (30)^99 + ... (90)^99] All the terms in a single group have the same last digit (i.e. last digits of 199 + 1199 + ... + 9199 are same, is 1, & similarly for the other groups). Also, there are 10 terms in each group except for the last one. Therefore the last digit of the sum of terms in first 9 groups is 0. (as whatever be the last digit, we have to multiply it by 10) And the last digit of the sum of the terms in the group 10 is obviously 0. Hence, the last digit of the series is 0.

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 Subject Author Date solution Praneeth Yalavarthi 2007-07-09 08:41:41 re(2): last digit Ady TZIDON 2004-01-31 13:23:30 re(2): Last two digits Nick Hobson 2004-01-31 07:57:31 re: Last two digits Richard 2004-01-30 22:09:15 Last two digits Nick Hobson 2004-01-30 15:22:31 re: last digit e.g. 2004-01-27 09:16:50 last digit Purna 2004-01-24 05:08:27 No Subject Purna 2004-01-24 05:04:45 No Subject Purna 2004-01-24 05:04:34 re(4): solution Ady TZIDON 2004-01-23 10:41:49 re(3): solution Charlie 2004-01-23 10:38:36 re(3): solution --- Gosh!!!.. Ady TZIDON 2004-01-23 10:38:04 re(2): solution SilverKnight 2004-01-23 10:36:32 re(2): solution --- Gosh!!!.. Charlie 2004-01-23 10:22:38 re(2): solution Ady TZIDON 2004-01-23 10:21:36 re: solution Charlie 2004-01-23 10:07:55 re: solution Ady TZIDON 2004-01-23 10:03:35 SOLUTION Ady TZIDON 2004-01-23 09:57:59 solution SilverKnight 2004-01-23 09:47:03

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