Last Digit (Posted on 2004-01-23)

	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.

	Subject	Author	Date
	Explanation to Puzzle Answer	K Sengupta	2022-09-07 02:58:35
	Puzzle Answer	K Sengupta	2022-08-31 22:50:03
	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