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If p and q are both positive integers such that p/9 + q/10 i

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If p and q are both positive integers such that p/9 + q/10 i [#permalink] New post 01 Apr 2018, 12:56
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If p and q are both positive integers such that \(\frac{p}{9} + \frac{q}{10}\) is also an integer, then which one of the following numbers could p equal?

(A) 3
(B) 4
(C) 9
(D) 11
(E) 19
[Reveal] Spoiler: OA

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Re: If p and q are both positive integers such that p/9 + q/10 i [#permalink] New post 01 Apr 2018, 21:22
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Let's solve this 2 different ways: the sum and difference rule and just logic.

METHOD 1:
If we give these two fractions a common denominator, they can be rewritten as:

10p+9q
90

Since this must equal an integer, 10p+9q has to total to 90, or 180, or 270, or some multiple of 90. We can make an equation therefore like this:

10p + 9q = 90x (where x is some integer just saying what multiple of 90 we're making.)

Now there's a pretty useful rule called the Sum and Difference Rule, which simply stated is this: if you have an equation in which 3 or more integer terms are being added or subtracted, and all but one obviously have a common factor, then the last one also must have that same factor. Think of it like this: if you buy several cartons of a dozen eggs, and your friend also buys a few cartons of a dozen eggs, than together, you must have some multiple of 12 eggs, right? Or if you bought several of these cartons of a dozen eggs, and your friend bought some, and you noticed that together you had 120 eggs, then since you have 10 dozen eggs, your friend couldn't have bought just a pillowcase full of eggs, right? Your friend also had to have bought exactly some multiple of 12 eggs as well. Anyway, back to our problem.

Since we see that in 10p + 9q = 90x, the two terms on the right both have 9 as a common factor, we know that the first term also has 9. But since 10 doesn't have 9 as a factor, the 9 must be a factor of p! Thus, since C is the only answer choice with 9 as a factor, it's the correct answer. Note that if we had an answer of 18 or 99, those could also be correct answers. (BONUS QUESTION: What do we know about q? Answer: by the same logic, q has to have 10 as a factor.)

METHOD TWO:
If you look at answer choice A and plug it in for p, you'll see that we have 1/3 + q/10, which is supposed to equal an integer. But no matter what you make q, you'll never be able to make it something that'll add cleanly with 1/3 to make an integer.

Similarly, if you look at answer choice B, getting us 4/9 + q/10, again you can tell basically by instinct that you'll never get an integer for q that will add cleanly to 4/9. The decimal version of 4/9 is .4444 repeating forever, and if you choose any integer for q, and divide it by 10, you'll always get a decimal that terminates. Since a terminating decimal and a non-terminating decimal are never going to add to make an integer, we can rule out B. D and E, similarly, would make non-terminating decimals as well.

However, if p is 9, it will cancel out the denominator of 9, giving us 1. The second variable, q, could obviously be 10 or some other multiple of 10, giving us another integer. Thus, C is the answer.
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Re: If p and q are both positive integers such that p/9 + q/10 i [#permalink] New post 21 May 2018, 22:43
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Assuming that Q will be a multiple of 10. For P it has to be a multiple of 9 otherwise it will generate a decimal number.

when a decimal number is added to an integer it will result in a decimal number.
A,B,D,E are eliminated hence the option is C.
Re: If p and q are both positive integers such that p/9 + q/10 i   [#permalink] 21 May 2018, 22:43
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If p and q are both positive integers such that p/9 + q/10 i

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