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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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Solution Explanation to Puzzle Answer Comment 21 of 21 |
(In reply to Puzzle Answer by K Sengupta)

The given expression is equal to:

{1^99+2^99+3^99+........+48^99+49^99} +50^99+
{(100-49)^99+(100-48)^99+.........+(100-3)^99+(100-2)^99+(100-1)^99}
 Reducing the terms and rearranging them mod 100, we have:
[{1^99+(-1)^99}+{2^99+(-2)^99}+{3^99+(-3)^99}+...........+{48^99+(-48)^99}
+{49^99+(-49)^99} +50^99] (mod 100)
== 50^99 (mod 100) 
==2500 * 50^97 (mod 100)
== 0(mod 100)
Accordingly, the last two digits of the given expression is 00
Consequently,  the last digit of the given expression is 0

Edited on September 7, 2022, 2:59 am
  Posted by K Sengupta on 2022-09-07 02:58:35

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