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If A and D are positive integers, what is the remainder when 7^(40A) +

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If A and D are positive integers, what is the remainder when 7^(40A) + [#permalink]

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New post 27 Apr 2016, 00:10
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Kudos [?]: 128932 [1], given: 12183

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Re: If A and D are positive integers, what is the remainder when 7^(40A) + [#permalink]

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New post 27 Apr 2016, 01:58
[quote="Bunuel"]If A and D are positive integers, what is the remainder when 7^(40A) + D is divided by 5?

(1) A = 5
(2) D = 4

7^200 ends in 1 ...so if D is 4 then rem=0 else reminder = something else
pattern of 7 ends in 7,9,3,1

7+4=11 reminder is different
9+4=13....

Combination yes possible C
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Re: If A and D are positive integers, what is the remainder when 7^(40A) + [#permalink]

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New post 27 Apr 2016, 02:02
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40A = 4*10*A --> Multiple of 4

7^40A will always have a units digit of 1 as cyclicity of 7^x is 7, 9, 3, 1.

A digit will be completely divisible by 5 if the units digit is 5 or 0.

St1: A = 5 --> Not Sufficient as we already know that 7^40A will have units digit as 1

St2: D = 4 --> Units digit(7^40A) + D = 1 + 4 = 5
The term is completely divisible by 5 and hence remainder is 0.
Sufficient

Answer: B

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Re: If A and D are positive integers, what is the remainder when 7^(40A) + [#permalink]

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Re: If A and D are positive integers, what is the remainder when 7^(40A) +   [#permalink] 15 Aug 2017, 05:41
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