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a, b, and c are positive integers. If a, b, and c are assembled into

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a, b, and c are positive integers. If a, b, and c are assembled into [#permalink]

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New post 22 May 2016, 13:36
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Question Stats:

24% (01:26) correct 76% (00:39) wrong based on 62 sessions

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a, b, and c are positive integers. If a, b, and c are assembled into the six-digit number abcabc, which one of the following must be a factor of abcabc?

(A) 16
(B) 13
(C) 5
(D) 3
(E) none of the above
[Reveal] Spoiler: OA

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Last edited by Bunuel on 23 May 2017, 04:20, edited 2 times in total.
Edited the OA.

Kudos [?]: 135629 [0], given: 12705

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Re: a, b, and c are positive integers. If a, b, and c are assembled into [#permalink]

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New post 23 May 2016, 02:19
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abcabc = \(10^5a + 10^4b + 10^3c + 10^2a + 10b + c\)
= \((10^5 + 10^2)a + (10^4 + 10)b + (10^3 + 1)c\)
= \(10^2(10^3 + 1)a + 10(10^3 + 1)b + (10^3 + 1)c\)
= \((10^3 + 1)(10^2a + 10b + c)\)
= \(1001 (100a + 10b + c)\)

1001 = 7 * 11 * 13

Hence the integer abcabc must be divisible by 7, 11 and 13

Answer: B

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Re: a, b, and c are positive integers. If a, b, and c are assembled into [#permalink]

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New post 23 May 2016, 10:39
Bunuel wrote:
a, b, and c are positive integers. If a, b, and c are assembled into the six-digit number abcabc, which one of the following must be a factor of abcabc?

(A) 16
(B) 13
(C) 5
(D) 3
(E) none of the above


Plug in some values and check -

abcabc = 123123

Not divisible by 16 and 5

let abcabc = 125125

Not divisible by 3

Only option (B) and (E) is left in both the cases...

Check once more to marke (B) as correct answer

let abcabc = 135135

Again divisible by 13

So, mark answer as (B) 13

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Re: a, b, and c are positive integers. If a, b, and c are assembled into [#permalink]

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New post 03 Dec 2017, 10:28
it is clearly visible that abcabc = 1001*abc
now checking with options, we get that 1001 is divisible by 13. hence B is the answer.
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Kudos [?]: 23 [0], given: 24

Re: a, b, and c are positive integers. If a, b, and c are assembled into   [#permalink] 03 Dec 2017, 10:28
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