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If j and k are positive integers, what is the remainder when (8 * 10^k

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If j and k are positive integers, what is the remainder when (8 * 10^k  [#permalink]

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New post 15 Jan 2015, 06:21
1
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A
B
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  25% (medium)

Question Stats:

78% (01:13) correct 22% (01:16) wrong based on 210 sessions

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Re: If j and k are positive integers, what is the remainder when (8 * 10^k  [#permalink]

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New post 15 Jan 2015, 09:11
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Bunuel wrote:
If j and k are positive integers, what is the remainder when (8 * 10^k) + j is divided by 9?

(1) k = 13
(2) j = 1

Kudos for a correct solution.


reminder of \(8(10^k)/9\) is always 8 no matter what positive integer value k assumes. If we know what is j, we are good to go.

1) Redundant.
2) Sufficient. the expression is a multiple of 9. Reminder is zero.

Answer B.
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Re: If j and k are positive integers, what is the remainder when (8 * 10^k  [#permalink]

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New post 15 Jan 2015, 11:10
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If j and k are positive integers, what is the remainder when (8 * \(10^k\)) + j is divided by 9?

(1) k = 13 (adding powers to K will not give the ans + J is required) so 1st statement is insufficient ,Therefore A,D out
(2) j = 1 ( Value of J is sufficient to reach at one ans whether value of k is mentioned or not) hence sufficient

Hence B! ans


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Re: If j and k are positive integers, what is the remainder when (8 * 10^k  [#permalink]

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New post 15 Jan 2015, 17:22
1
Bunuel wrote:
If j and k are positive integers, what is the remainder when (8 * 10^k) + j is divided by 9?

(1) k = 13
(2) j = 1

Kudos for a correct solution.


1. Insufficient: Divisibility of this expression with k=13 will depend on what is added to (8 * 10^k) as the integer j.
2. Sufficient: Irrespective of the value of k (which will only increase the multiple of 10), adding j to 8* any multiple of ten always gives a remainder of zero, that is, is divisible by 9.

Answer B.
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Re: If j and k are positive integers, what is the remainder when (8 * 10^k  [#permalink]

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New post 09 Dec 2017, 04:34
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Re: If j and k are positive integers, what is the remainder when (8 * 10^k &nbs [#permalink] 09 Dec 2017, 04:34
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