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

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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 5?

(1) n has a value of either 6 or 12.
(2) n is a multiple of 6.
[Reveal] Spoiler: OA

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Re: If x and n are integers such that x = 1^1 + 2^2 + 3^3 + . . . + n^n, [#permalink]

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New post 09 Jan 2017, 06:14
Bunuel wrote:
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 5?

(1) n has a value of either 6 or 12.
(2) n is a multiple of 6.


(1)
n=6

\(\frac{(1^1 + 2^2 + 3^3 + 4^4 + 5^5 + 6^6)}{5} = \frac{(1 + (-1) + 2 + (-1)^4 + 0 + 1^6)}{5} = \frac{4}{5}\) remainder 4

n=12

\(7^7 + 8^8 + 9^9 + 10^{10} + 11^{11} + 12^{12}\)

Last digits: \(\frac{(3 + 6 + 9 + 0 + 1 + 6)}{5} = \frac{5}{5}\) remainder 0 ---->Total remainder n=12 = 4 + 0 = 4

Sufficient

(2) We'll take next multiple of 6 - 18

\(13^{13} + 14^{14} + 15^{15} + 16^{16} + 17^{17} + 18^{18}\)

Last digits - 3 + 6 + 5 + 6 + 7 + 4 = 31 remainder 1

Total remainder n=18 4 + 1 = 5 ---> remainder 0.

We have remainders 4 and 0. Insufficient.

Unswer A
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Re: If x and n are integers such that x = 1^1 + 2^2 + 3^3 + . . . + n^n, [#permalink]

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New post 10 Jan 2017, 18:08
This one is E for me.

Way I went about this: count just the units digits of 1^1, 2^2...all the way up to 12^12 since statement 1 specifics n is either 6 or 12. Taking the first statement, if n=6, then we have:

1+4+7+6+5+6.....adds up to 29, remainder is 4.

On the other hand, if n=12, then we have:

1+4+7+6+5+6+3+6+9+0+1+2....adds up to 50. Remainder when divided by 5 is 0. Insufficient.

Statement 2: n is a multiple of 6. We just proved 6 and 12 will yield different remainders when divided by 5. Insufficient.

Both together: we don't learn anything new, 6 and 12 still fit and still yield different remainders.

E.
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Re: If x and n are integers such that x = 1^1 + 2^2 + 3^3 + . . . + n^n, [#permalink]

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New post 10 Jan 2017, 21:31
mcrimmin wrote:
This one is E for me.

Way I went about this: count just the units digits of 1^1, 2^2...all the way up to 12^12 since statement 1 specifics n is either 6 or 12. Taking the first statement, if n=6, then we have:

1+4+7+6+5+6.....adds up to 29, remainder is 4.

On the other hand, if n=12, then we have:

1+4+7+6+5+6+3+6+9+0+1+2....adds up to 50. Remainder when divided by 5 is 0. Insufficient.

Statement 2: n is a multiple of 6. We just proved 6 and 12 will yield different remainders when divided by 5. Insufficient.

Both together: we don't learn anything new, 6 and 12 still fit and still yield different remainders.

E.


Hi

12^{12} ---> 2^{12}

exponent 12 is a multiple of 4, hence last digit is 6 not 2.

The sum of digits when n=12 will add up to 54 with remainder 4.
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Re: If x and n are integers such that x = 1^1 + 2^2 + 3^3 + . . . + n^n, [#permalink]

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New post 01 Mar 2017, 23:23
Bunuel wrote:
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 5?

(1) n has a value of either 6 or 12.
(2) n is a multiple of 6.

Bunuel
please check the answer provided..I'm getting A as well.
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If x and n are integers such that x = 1^1 + 2^2 + 3^3 + . . . + n^n, [#permalink]

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New post 01 Mar 2017, 23:40
if n=6, it yields 29...remainder /5 is 4, n=12 yields 54...rem/5 is again..4....but when n=18.it is 85/rem5= 0
Thus ans A
If x and n are integers such that x = 1^1 + 2^2 + 3^3 + . . . + n^n,   [#permalink] 01 Mar 2017, 23:40
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