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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 Updated on: 23 May 2017, 05:20
00:00
A
B
C
D
E

Difficulty:

  95% (hard)

Question Stats:

40% (01:50) correct 60% (01:20) wrong based on 155 sessions

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Originally posted by Bunuel on 22 May 2016, 14:36.
Last edited by Bunuel on 23 May 2017, 05:20, edited 2 times in total.
Edited the OA.
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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, 03:19
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1
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, 11:39
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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, 11:28
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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 27 Nov 2018, 19:02
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Top Contributor
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


Let's think of abc as a number (NOT a product).
So, for example, if a = 1, b = 2 and c = 5, then abc = 125, and abcabc = 125125

In this regards, abcabc = abc000 + abc
= abc(1000 + 1)
= abc(1001)
= abc(7)(11)(13)

So, abcabc must be divisible by 7, 11 and 13

Answer: B

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Brent
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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 27 Jun 2018, 10:32
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

Remember that
any three digit number when multiplied by 1001 , the result is the same three digit number side by side
abc X 1001 = abcabc

or, abc X 13 X 11 X 7 = abcabc

which is surely divisible by 13, 11 and 7

answer is B :-D
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a, b, and c are positive integers. If a, b, and c are assembled into   [#permalink] 27 Jun 2018, 10:32
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