If x and n are integers such that x = 1^1 + 2^2 + 3^3 + . . . + n^n,

Question

If x and n are integers such that x = 1^1 + 2^2 + 3^3 + . . . + n^n, what is the remainder when x is divided by 10?

(1) n is a multiple of 6 
(2) n = 6

Vinayak Shenoy · Answer

IMO B
Given x=1^1+ 2^2+ 3^3+4^4+.....+n^n

from statement 1
n can be 6, 12, 18
when n=6, x= 1+4+27+256+3125+46656=(taking only units digit we have) 1+4+7+6+5+6=29 Remainder=9 when divided by 10
when n=12, x= 1+4+7+6+5+6+3+6+1+0+1+6=46 Remainder=6 when divided by 10
hence insufficient

From statement 2
n=6 hence remainder=9
Sufficient.
Hence B

arpitkansal · Answer

Bunuel   chetan2u  can you explain how to approach this question?

chetan2u · Answer

Hi..

remainder when divisible by 10 means the units digit.. 
Now x= 1^1+2^2+3^3+4^4+5^5+....=1^1+2^2+3^3+4^4+5^1+6^2+... 
We can write this since the units digit repeat after every 4 powers...

Let us see the statements

1) n is a multiple of 6..
1st case...  1^1+2^2+3^3+4^4+5^1+6^2=1+4+7+6+5+6=29 
2nd case... 1^1+2^2+3^3+4^4+5^1+6^2+7^3+8^4+9^1+10^2+11^3+12^4=29+3+6+9+0+1+6=29+15 
So different units digit..

2) n is 6..
No need to check. You can always count when you know the variable
Sufficient

B

Sagar360 · Answer

x = 1^1 + 2^2 + 3^3 + . . . + n^n = {n(n+1)(2n+1)}/6

The above summation formula makes it easier

bumpbot · Answer

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If x and n are integers such that x = 1^1 + 2^2 + 3^3 + . . . + n^n,

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