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ABA represents a three-digit number having digits

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Source: — Data Sufficiency |

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by Jay@ManhattanReview » Tue Jun 13, 2017 11:29 pm
stevecultt wrote:ABA represents a three-digit number having digits A and B; BC3 represents another three digit having digits B, C and 3. What is the value of B?

(1) ABA x 3 = BC3
(2) A, B, and C are distinct non-zero digits.

OA E
Statement 1: ABA x 3 = BC3

Since the unit digit of 'ABA x 3' is determined by Ax3, which is '3' as the unit digit of BC3 is 3, A = 1.

Thus, we have, 1B1 x 3 = BC3

=> 3(100 + 10B + 1) = 100B + 10C + 3

=> 303 + 30B = 100B + 10C + 3

=> 30 = 7B + C

The maximum value of C would be 9. Thus, the minimum value of B = (30 - 9)/7 =3

The other value of B could be 4, rendering C = 30 - 7*4 = 2.

Thus, either A = 1, B = 3, and C = 9 OR A = 1, B = 4, and C = 2. No unique value of B. Insufficiet.

Statement 2: A, B, and C are distinct non-zero digits.

Clearly, insufficent.

Statement 1 & 2 combined:

Even both the statement together is not sufficient as either A = 1, B = 3, and C = 9 OR A = 1, B = 4, and C = 2. No unique value of B.

The correct answer: E

Hope this helps!

Relevant book: Manhattan Review GMAT Data Sufficiency Guide

-Jay
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by [email protected] » Wed Jun 14, 2017 5:26 pm
Hi stevecultt,

This DS question is built around a few Number Property rules, but even if you don't spot those pattern you can still get the correct answer with a little 'brute force' arithmetic.

We're told that A, B and C are DIGITS in the three-digit numbers ABA and BC3. This means that each of those variables can only be the numbers 0-9 inclusive. We're asked for the value of B.

1 )ABA x 3 = BC3

There are only 10 possible values for A - and Fact 1 gives us an equation that tells us that when we multiply A by 3, we get a number that ENDS in a 3. Could A = 5? NO - because multiplying a number that ends in 5 by 3 will gives us a number that ends in a 5. How long would it take you to multiply each possible digit by 3 and see what happens? You'll find that there's just ONE possibility that 'fits'... A = 1. With that value 'locked in', we have...

1B1 x 3 = BC3

From here, we can take the same approach to figure out the possible value(s) of B. Could B = 1? NO - because 111 x 3 = 333 (not 1C3). How long would it take to work through the other possibilities (hint: you won't have to work for very long). You'll find that B = 3 and B = 4 are both possibilities.

131 x 3 = 393
141 x 3 = 423
Fact 1 is INSUFFICIENT.

2) A, B, and C are distinct non-zero digits.

Fact 2 lets us know that the three variables are all different from one another, but there's still no way to determine the exact value of B.
Fact 2 is INSUFFICIENT.

Combined, we still end up with two possible outcomes (A=1, B=3, C=9 and A=1, B=4, C=3).
Combined, INSUFFICIENT.

Final Answer: E

GMAT assassins aren't born, they're made,
Rich
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