Find the remainder when 1 1 + 2 2 + 3 3 + 4 4 + ⋯ + 5 0 5 0 is divided by 8.
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
Exactly the same method sir! For those who didn't understand the comment, k k ≡ k ( m o d 8 ) for all odd k ,consider this,let k = 4 n ± 1 ,then k k = k × ( 4 n ± 1 ) a n e v e n n u m b e r ,now consider this, ( 4 n ± 1 ) 2 = 1 6 n 2 + 1 ± 8 n ≡ 1 ( m o d 8 ) .
Log in to reply
@Adarsh Kumar
Is there any alternate solution, using Euler's Totient Function?
Neat , short and elegant .Nice !
Did same...
If I divide each odd number by 8, getting remainder 1, 3, 5 and 7. There are 6 times of 1,3,5 and 7 and only one 4 I got. In this I got 100. So i got 4 as answer. Please tell me where I went wrong
Problem Loading...
Note Loading...
Set Loading...
(All congruences will be mod 8.) We have 2 2 = 4 and k k ≡ 0 for even k ≥ 4 . For odd k we have k 2 ≡ 1 so k k ≡ k . Thus our sum is 4 + 1 + 3 + 5 + . . . + 4 7 + 4 9 = 4 + 2 5 2 ≡ 5