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Bunuel
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If x and n are integers such that x = 1^1 + 2^2 + 3^3 + . . . + n^n, 03 Nov 2016, 02:47
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55% (hard)

Question Stats:

64% (02:07) correct 36% (02:00) wrong based on 192 sessions

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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
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B
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If x and n are integers such that x = 1^1 + 2^2 + 3^3 + . . . + n^n, 03 Nov 2016, 03:07
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Bunuel
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
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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
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If x and n are integers such that x = 1^1 + 2^2 + 3^3 + . . . + n^n, 12 Feb 2019, 09:42
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Bunuel chetan2u can you explain how to approach this question?
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If x and n are integers such that x = 1^1 + 2^2 + 3^3 + . . . + n^n, 12 Feb 2019, 10:00
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arpitkansal
Bunuel chetan2u can you explain how to approach this question?
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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
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If x and n are integers such that x = 1^1 + 2^2 + 3^3 + . . . + n^n, 14 Mar 2019, 08:55
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x = 1^1 + 2^2 + 3^3 + . . . + n^n = {n(n+1)(2n+1)}/6

The above summation formula makes it easier
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If x and n are integers such that x = 1^1 + 2^2 + 3^3 + . . . + n^n, 21 Nov 2023, 16:13
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