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Bunuel
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The sum of the last digits of the numbers of the form 2^(2n + 1), for
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15 Jul 2021, 04:30
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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
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The sum of the last digits of the numbers of the form 2^(2n + 1), for
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 Updated on: 16 Jul 2021, 05:59
Originally posted by MathRevolution on 15 Jul 2021, 07:21.
Last edited by MathRevolution on 16 Jul 2021, 05:59, edited 1 time in total.
MathRevolution
=> \(2^{2n + 1}\) 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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The sum of the last digits of the numbers of the form 2^(2n + 1), for
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15 Jul 2021, 08:01
MathRevolution
=> \(2^{2n + 1} at n = 0 => 2\)

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

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

This forms a G.P. with a = 2 and r = 4

Sum of G.P. with n = 5, a = 2 and r = 4 : \(\frac{a(r^n - 1) }{ r - 1} = \frac{2(4^5 - 1) }{ 4 - 1} = \frac{2(1023) }{ 3} = 682\)

682 divided by 7 => 682 = 97 * 7 + 3

Remainder = 3

Answer C
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Dear MathRevolution team,

In question, it is mentioned last digit of form 2^(2n+1)
Means when n=0 , it will be 2
When n=1, it will be 8
When n=2, it will be 2
When n=3, it will be 8
When n=4, it will be 2
Sum will be 22
Divided by 7 will give remainder 1.

Can you please correct

Regards
Rajat

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The sum of the last digits of the numbers of the form 2^(2n + 1), for
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15 Jul 2021, 08:05
Bunuel
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
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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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The sum of the last digits of the numbers of the form 2^(2n + 1), for
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16 Jul 2021, 05:56
MathRevolution
=> \(2^{2n + 1}\) 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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The sum of the last digits of the numbers of the form 2^(2n + 1), for
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16 Jul 2021, 05:57
ueh55406
Bunuel
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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The sum of the last digits of the numbers of the form 2^(2n + 1), for
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