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The sum of the last digits of the numbers of the form 2^(2n + 1), for [#permalink]
Bunuel wrote:
The sum of the last digits of the numbers of the form \(2^{2n+1}\), for n = 0, 1, 2, 3 and 4, when divided by 7 gives a remainder

(A) 1
(B) 2
(C) 3
(D) 4
(E) 5


Sum of the LAST digits of the number of the form: \(2^{2n+1}\); for n = 0, 1, 2, 3 and 4,

number= 2,8,32,128,512 >> sum of last digits = 2+8+2+8+2 = 22

22 / 7; remainder is 1.

A, IMO.

MathRevolution, I don't think the question means sum of the numbers itself, that would be too obvious, even for a 500 lvl question.
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Re: The sum of the last digits of the numbers of the form 2^(2n + 1), for [#permalink]
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MathRevolution wrote:
=> \(2^{2n + 1}\) at n = 0 => 2

=> \(2^{2n + 1}\) at n = 1 => 8

=> \(2^{2n + 1}\) at n = 2 => 32

=> \(2^{2n + 1}\) at n = 3 => 128

=> \(2^{2n + 1}\) at n = 4 => 512

Sum of all last terms: 2 + 8 + 2 + 8 + 2 = 22

22 divided by 7 => 22 = 3 * 7 + 1

Remainder = 1

Answer A
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Re: The sum of the last digits of the numbers of the form 2^(2n + 1), for [#permalink]
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ueh55406 wrote:
Bunuel wrote:
The sum of the last digits of the numbers of the form \(2^{2n+1}\), for n = 0, 1, 2, 3 and 4, when divided by 7 gives a remainder

(A) 1
(B) 2
(C) 3
(D) 4
(E) 5


Sum of the LAST digits of the number of the form: \(2^{2n+1}\); for n = 0, 1, 2, 3 and 4,

number= 2,8,32,128,512 >> sum of last digits = 2+8+2+8+2 = 22

22 / 7; remainder is 1.



Thank you so much.

I did edit the solution.

Thanks again.
A, IMO.

MathRevolution, I don't think the question means sum of the numbers itself, that would be too obvious, even for a 500 lvl question.
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Re: The sum of the last digits of the numbers of the form 2^(2n + 1), for [#permalink]
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