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# Find all possible values of X

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Find all possible values of X [#permalink]

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08 Jan 2012, 20:02
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X is a 3-digit number with all digits being distinct.

One can crate 9 2-digit numbers using the digits of X.

X is a sum of all 9 numbers.

Find values of X.

Last edited by adgir on 09 Jan 2012, 13:08, edited 3 times in total.
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09 Jan 2012, 12:11
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HI, there. I'm happy to help with this.

This is a very cool math problem --- I had fun with it. Of course, it is way way harder than anything asked on a GMAT Math section.

So, let's say that X has the digits A, B, C, where A & B & C are all unequal. Thus, x = "ABC" = A100 + B10 + C

(I'm using quote marks for digit form, to distinguish it from an arithmetic expression.)

The nine two digit numbers are

"AA" = A10 + A
"AB" = A10 + B
"AC" = A10 + C

"BA" = B10 + A
"BB" = B10 + B
"BC" = B10 + C

"CA" = C10 + A
"CB" = C10 + B
"CC" = C10 + C

The sum of these is 33A + 33B + 33C = 33(A + B + C)

If X equals this sum, then X = 33(A + B + C) ----> X is a multiple of 33, which means that X would be divisible by 3.

If X is divisible by 3, then the sum of it's digits (A + B + C) is also divisible by 3, and X = 33*(multiple of 3), so X is divisible by 9.

If X is divisible by 9, then the sum of the digits (A + B + C) is also divisible by 9.

So if X equals the sum of those nine two-digit numbers, then it must be a product of 33 and a multiple of 9. There are only three numbers like that less than 1000:

9*33 = 297
18*33 = 594
27*33 = 891

All three of those three-digit numbers have that add up to 18. So, the only one that will equal the sum as required is 594.

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

I believe that x = 594 is the only possibility for this condition.

Please let me know if you have any questions on this.

Mike
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Mike McGarry
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09 Jan 2012, 12:47
Thanks! Much appreciated!

Yes, it is beyond GMAT (at least I hope so!:) Just really got stuck on it and it was bugging me, so I couldn't let it go.
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09 Jan 2012, 20:43
mikemcgarry wrote:
If X equals this sum, then X = 33(A + B + C) ----> X is a multiple of 33, which means that X would be divisible by 3.

If X is divisible by 3, then the sum of it's digits (A + B + C) is also divisible by 3, and X = 33*(multiple of 3), so X is divisible by 9.

Hi, how can you say that here if x is divisible by 3 then (A+B+C) is also divisible by 3???
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Re: Find all possible values of X [#permalink]

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10 Jan 2012, 05:54
Thanks for the detailed explanation!
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Re: Find all possible values of X [#permalink]

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10 Jan 2012, 11:44
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
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Re: Find all possible values of X [#permalink]

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10 Jan 2012, 20:03
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???
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Re: Find all possible values of X [#permalink]

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11 Jan 2012, 09:15
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, http://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
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Mike McGarry
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Re: Find all possible values of X   [#permalink] 11 Jan 2012, 09:15
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