Is the three-digit positive integer ABC divisible by 7?

Question

(1) The two-digit positive integer AB minus 2*c is divisible by 7.
(2) 2*A + 3*B + C is divisible by 7.

Is the 3-digit number abc divisible by 7? (a, b, and c are the respective digits of the 3-digit number)

(1): The 2-digit number ab minus (c * 2) is divisible by 7.
(2): 2*a + 3*b + c is divisible by 7.

chetan2u · Answer

Bunuel , you have missed out the value of the two-digit positive integer in variables. I think it should have been AB, subtract 2*C

Bunuel · Answer

_______________
Fixed. Thank you!

IanStewart · Answer

These are somewhat famous divisibility tests for 7, if you learn about those sorts of things (they're never useful on the GMAT, so there's no reason to), and the answer here is D.

One would normally prove these tests work using modular arithmetic, which is beyond the scope of the GMAT. We don't strictly need to use it:.

From Statement 1, 10A + B - 2C is divisible by 7. So it equals some multiple of 7, which we can write as 7q. Then if we multiply by 10 on both sides, and rewrite things to get 100A + 10B + C (which is the three digit number ABC) we find:

10A + B - 2C = 7q
100A + 10b - 20C = 70q
100A + 10B + C - 21C = 70q
100A + 10B + C = 70q + 21C

and since we're adding two multiples of 7 on the right side, the right side is divisible by 7, so the left side must be also, since it's the same number as the right side. So the three-digit number ABC is divisible by 7 and Statement 1 is sufficient.

Using Statement 2, if 2A + 3B + C is divisible by 7, then we must get another multiple of 7 if we add 98A, because 98A is divisible by 7 (since 98 = 7*14). We must also get a multiple of 7 if we add 7B. So if 2A + 3B + C is divisible by 7, so is 2A + 98A + 3B + 7B + C = 100A + 10B + C, but that's the three-digit number ABC. So Statement 2 is sufficient.

But I wouldn't expect a question on the real GMAT to expect a test taker to complete that kind of analysis.

Tonkotsu · Answer

Thanks to  IanStewart  these questions finally makes sense. 
Too bad that they may not be on the gmat, since they're a lot easier than I thought.

Adding 98A was a bit confusing for me on statement 2, since I don't think I could readily reproduce and think of such an approach.
However, I finally realized, that as long as I can match the equation  100A + 10B + C = DQ , I could probably always solve it.

2A + 3B + C = 7Q

Multiply both sides by 50, to at least make 100A.

100A + 150B + 50C = 350Q

Shuffle

100A + 10 B + C = 350Q -140B -49C

RHS is divisible by 7, therefore sufficient!

IanStewart · Answer

Nice solution! I think I like it more than the one I posted.

bumpbot · Answer

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Is the three-digit positive integer ABC divisible by 7?

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