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

Question

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

PareshGmat · Accepted Answer

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

chetan2u · Answer

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

Kurtosis · Answer

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

Abhishek009 · Answer

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

shashankism · Answer

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

GMATBusters · Answer

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.

BrentGMATPrepNow · Answer

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

Kritisood · Answer

can we generalize that a 6 digit number is always divisible by 7, 11, and 13?

chetan2u · Answer

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

KarishmaB · Answer

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)

bumpbot · Answer

Automated notice from GMAT Club BumpBot: 

A member just gave Kudos to this thread, showing it’s still useful. I’ve bumped it to the top so more people can benefit. Feel free to add your own questions or solutions.

 This post was generated automatically.

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

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions

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

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards