Find all possible values of X

Hi, how can you say that here if x is divisible by 3 then (A+B+C) is also divisible by 3???

Thanks for the detailed explanation!

This is a reply to subhajeet's question: "how can you say that here if x is divisible by 3 then (A+B+C) is also divisible by 3???"

This is what I would call a BIG idea on GMAT math.

If a number if divisible by three, then the sum of it's digits is also divisible by three.

For example, 15 is divisible by three, and the sum of it's digits, 1 + 5 = 6, is also divisible by three.

This is a great trick to know with larger numbers. Is 2012 divisible by three? Well, the sum of its digits, 2 + 0 + 1 + 2 = 5, and that's not divisible by three, so that means, 2012 can't be divisible by three either. By contrast, with 2013, we have a sum of 2 + 0 + 1 + 3 = 6, which is divisible by three, so that means 2013 must be divisible by three.

This is a great trick, and the GMAT expects you to know it.

There is a corresponding trick with 9. If a number if divisible by nine, then the sum of it's digits is also divisible by nine. Same idea --- the sums from 2012 and 2013 are not divisible by 9; the next year that will be divisible by 9 is 2016, because 2 + 0 + 1 + 6 = 9.

I can't stress enough: both of these are in the category of fundamental number properties that the GMAT simply expects you to know.

Please let me know if you have any further questions.

Mike

HI Mike, I agree with your above post that every number has its own property. But my question is that if X = 33(A+B+C) => X is divisible by both 3 and 11. So why are we not taking into account that also???

User subhajeet wrote: "HI Mike, I agree with your above post that every number has its own property. But my question is that if X = 33(A+B+C) => X is divisible by both 3 and 11. So why are we not taking into account that also???"

That's a great question. If X - 33*(something), then X we absolutely know that X must be divisible by 3 and 11. I used the divisibility by 3 in my solution, but not the divisibility by 11. Why?

There are a couple ways to answer that question. One is: the GMAT definitely expects you to be able to look at a large number and tell whether it's divisible by three. That is one that I think I've seen at least on every full-length GMAT I've taken or studied. By contrast, divisibility by 11 -- yes, there is a relatively simple trick for that as well, but I have never seen a real GMAT question where the test expected we would know that. So, that's part of the answer: divisibility by three is highly GMAT-relevant, and divisibility by 11 isn't at all.

The deeper answer has to do with the peculiarity of mathematical problem-solving. The route I happened to see to the answer to the original question was

algebraic formulation --> divisibility by 3 ----> divisibility by 9 ---> multiples of 9 times 33 ----> few enough to check one-by-one

Is that the best way to answer this question? I don't know. It's the best of which I could conceive. As it happens, in this route, divisibility by three was vital, and divisibility by 11, while perfectly true, was completely irrelevant. I point out, though, it's not surprising that the GMAT-practice source would include a question in which divisibility by 3 plays a vital role in the solution, since that's an important concept in the GMAT Math.

Perhaps another way to say it: in doing math at this level, we need to distinguish between (a) what's mathematically allowed, mathematically legal, vs. (b) what's strategic: what will move me closer to the answer? Think about a very simple algebra equation: Given 2x + 5 = 13, solve for x. There are an infinity of mathematically legal steps we could take, but most of them would be nonstrategic. For example, I could begin by adding 317 to both sides of the equation: that is absolutely legal mathematically, but in terms of strategy it would be a completely daft move. Just because you can do something, just because something is mathematically legal, does not mean that it's strategic, does not mean that it's a move that will pay dividends in getting you that much closer to the answer. In this problem, it would have been perfectly mathematically legal to explore divisibility by 11 instead of divisibility by 3, but so far as I can tell, the former path doesn't really go anywhere, whereas, the latter path leads elegantly to the solution.

Of all the things that are mathematically legal, how do you determine what's most strategic? There's no short answer to that one. Nothing replaces experience with problem-solving. There is a classic, relatively dense text by the mathematician George Polya, How to Solve It, if you want to have a more theoretical introduction. (This pdf, https://furius.ca/cqfpub/doc/proofs/how-to.pdf, gives the gist of Polya's approach). But, fundamentally, math is not a spectator sport: you learn it, you become better at it, only by doing it.

I hope, at least to some extent, that answers your question. Please let me know if you have questions on anything I have said.

Mike

Amazing explanation mikemcgarry.
594 Kudos from my side....

By the way just to explain more about your statement "So, the only one that will equal the sum as required is 594."
This is because if u take 2 digit cases from all 3 possible values 297, 594 & 891. The sum of all 9 digits for either of the 3 values, will be always come to be 594.

What I mean is:-

Case 1 :-
X= 297

then,
22+29+27+99+92+97+77+72+79 = 594


Case 2:-
X = 594

then again,
55+59+54+95+99+94+45+49+44 = 594

similarly Case 3:-
X = 891

then,
88+89+81+98+99+91+18+19+11 = 594

Thus as you said, the only one that will equal the sum as required is 594.