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

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Bunuel
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a, b, and c are positive integers. If a, b, and c are assembled into t [#permalink] New post   27 Oct 2014, 07:22
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44% (01:55) correct 56% (01:43) wrong based on 313 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
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B
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PareshGmat
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Re: a, b, and c are positive integers. If a, b, and c are assembled into t [#permalink] New post   27 Oct 2014, 20:41
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Bunuel wrote:

Tough and Tricky questions: Factors.



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


Option A, C & E can be safely ignored as we can pick up some numbers which contradict the divisibility

247247 >> Cannot be divided by 16, 5 & 3

To check for 13, better to go the algebraic way:

\(abcabc = a * 10^5 + b * 10^4 + c * 10^3 + a * 10^2 + b * 10^1 + c * 10^0\)

\(= a10^2(1000+1) + b10(1000+1) + c(1000+1)\)

Check divisibility of 1001 by 13

\(\frac{1001}{13} = 77\)

Answer = B = 13
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Re: a, b, and c are positive integers. If a, b, and c are assembled into t [#permalink] New post   09 Apr 2016, 04:31
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Bunuel wrote:

Tough and Tricky questions: Factors.



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



Here the BEST approach would be dividing by abc..
we can be sure that abcabcwill get div by abc..
abcabc/abc = 1001..
what are factors of 1001 = 7*11*13..
Thus in given choices 13 is correct
ans B
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Re: a, b, and c are positive integers. If a, b, and c are assembled into t [#permalink] New post   23 May 2016, 03:19
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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 t [#permalink] 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 t [#permalink] New post   12 Aug 2017, 07:55
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Bunuel wrote:

Tough and Tricky questions: Factors.



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


abcabc = 10^5*a + 10^4*b + 10^3*c + 10^2 *a + 10 * b + c
= 1001 (100a + 10b + c)

Now 1001 = 13*7*11

So, 13 must be a factor of abcabc

Answer B
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Re: a, b, and c are positive integers. If a, b, and c are assembled into t [#permalink] New post   03 Dec 2017, 11:28
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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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BrentGMATPrepNow
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Re: a, b, and c are positive integers. If a, b, and c are assembled into t [#permalink] New post   27 Nov 2018, 19:02
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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


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

Cheers,
Brent
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Re: a, b, and c are positive integers. If a, b, and c are assembled into t [#permalink] New post   07 May 2020, 03:57
can we generalize that a 6 digit number is always divisible by 7, 11, and 13?
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Re: a, b, and c are positive integers. If a, b, and c are assembled into t [#permalink] New post   07 May 2020, 04:54
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Kritisood wrote:
can we generalize that a 6 digit number is always divisible by 7, 11, and 13?


No, you cannot generalise this for all 6 digit number.
But it is true for all 6-digit numbers where from left 1st and 4th, 2nd and 5th & 3rd and 6th digits are same.
OR of the form abc abc, so 102102 or 100100 or 476476 are all divisible by 7, 11 and 13
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Re: a, b, and c are positive integers. If a, b, and c are assembled into t [#permalink] New post   25 Oct 2023, 22:38
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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


I often tell my learners that my favourite number on GMAT is 1001.
Here is a post on WHY: https://anaprep.com/number-properties-t ... -paradigm/

It is a good idea to remember that 1001 = 7 * 11 * 13

abcabc = abc * 1001

Hence, abcabc must be a factor of 13.

Answer (B)
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Re: a, b, and c are positive integers. If a, b, and c are assembled into t [#permalink]
25 Oct 2023, 22:38
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