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multiple

by maihuna » Mon Apr 20, 2009 9:55 am
if x and y are positive integers such that 3x+7y is a multiple of 11 then which of the following will also be dividible by 11:

1. 4x+6y
2. x+y+4
3. 9x+4y
4. 4x -9y
5. 7x + 4y

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by moutar » Mon Apr 20, 2009 11:05 am
Choose values of x and y such that 3x + 7y is a multiple of 11.
x = 5, y = 1 works

Plug each of these into the answers and you'll find that:

1. 4x + 6y = 20 + 6 = 26 WRONG
2. x + y + 4 = 5 + 1 + 4 = 10 WRONG
3. 9x + 4y = 45 + 4 = 49 WRONG
4. 4x - 9y = 20 - 9 = 11
5. 7x + 4y = 35 + 4 = 39 WRONG

Therefore option 4
Last edited by moutar on Mon Apr 20, 2009 11:17 am, edited 1 time in total.

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by dumb.doofus » Mon Apr 20, 2009 11:12 am
Is there any simpler way other than plugging values?
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by Ian Stewart » Mon Apr 20, 2009 3:55 pm
dumb.doofus wrote:Is there any simpler way other than plugging values?
There are quite a few ways to see why this works - there's surely a faster way than what I've done below, but it's the end of a long day here!

_______

If 3x+7y is a multiple of 11, then 3x + 7y - 22y = 3x - 15y is a multiple of 11 (whenever we add or subtract two multiples of 11, we must get a multiple of 11). So 3x - 15y = 3(x - 5y) is a multiple of 11. Since 3 is not a multiple of 11, x - 5y must be a multiple of 11. We know now that 3x + 7y and x - 5y are multiples of 11; if we add them we must get another multiple of 11:

3x + 7y
x - 5y
4x + 2y

So 4x + 2y is a multiple of 11, and so is 4x + 2y - 11y = 4x - 9y.
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by maihuna » Tue Apr 21, 2009 6:29 am
Ian,
Can you please elaborate further steps on how to pick the magic numbers here in the two steps mentioned by you.
Regards,
maihuna

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by Ian Stewart » Tue Apr 21, 2009 7:48 am
maihuna wrote:Ian,
Can you please elaborate further steps on how to pick the magic numbers here in the two steps mentioned by you.
Regards,
maihuna
Well, there's not really any 'magic' involved, because if you do any similar sequence of steps to what I did above, you'll arrive at the answer. Of course, it helps a lot to know the correct answer in advance; picking numbers, as moutar did above, seems the best approach in test conditions.

I only wanted to show *why* the relationship is true, no matter what x and y are. The goal is to rewrite the expression in its simplest form (with the smallest possible numbers). Factoring seems the best way to do that. Looking at 3x + 7y, there is no common factor, but if you add 11x (to get 14x + 7y), or 11y (to get 3x + 18y), for example (there are many other options), you can then factor something out, and get smaller numbers in your expression. So, for example, since 14x + 7y is divisible by 11, and since 14x + 7y = 7(2x + y), then 2x+y must be divisible by 11, since 7 is not. Once you discover that 2x + y must be a multiple of 11, it's easier to build up all of the combinations of x and y that must be divisible by 11. So, for example, you could instead do:

3x + 7y is divisible by 11
14x + 7y is divisible by 11 (add 11x)
2x + y is divisible by 11 (factor out 7)
4x + 2y is divisible by 11 (multiply by 2)
4x - 9y is divisible by 11 (subtract 11y)

In any case, it doesn't seem much like any real GMAT question I've ever seen.
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