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If p and q are both positive integers such that p/9 + q/10 is also an

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If p and q are both positive integers such that p/9 + q/10 is also an  [#permalink]

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New post 26 Jul 2017, 03:38
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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



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If p and q are both positive integers such that p/9 + q/10 is also an  [#permalink]

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New post 26 Jul 2017, 04:48
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Inten21 wrote:
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



Source: GMAT NOVA MATH BIBLE
A KUDO = A Thank you. :-)


Let \(k\) be an integer such that \(\frac{p}{9} + \frac{q}{10} = k\).

Since all three numbers \(p\), \(q\), and \(k\) are integers, among the given options, the only case could be:
integer \((\frac{p}{9})\) + integer \((\frac{q}{10})\) = integer \((k)\)

The only value that fits the condition is (C). So, Ans - C.
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Re: If p and q are both positive integers such that p/9 + q/10 is also an  [#permalink]

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New post 26 Jul 2017, 09:51
TimeTraveller wrote:
Inten21 wrote:
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



Source: GMAT NOVA MATH BIBLE
A KUDO = A Thank you. :-)


Let \(k\) be an integer such that \(\frac{p}{9} + \frac{q}{10} = k\).

Since all three numbers \(p\), \(q\), and \(k\) are integers, among the given options, the only case could be:
integer \((\frac{p}{9})\) + integer \((\frac{q}{10})\) = integer \((k)\)

The only value that fits the condition is (C). So, Ans - C.


That's absolutely correct. Great work.
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If p and q are both positive integers such that p/9 + q/10 is also an  [#permalink]

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New post 26 Jul 2017, 10:14
Inten21 wrote:
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

Source: GMAT NOVA MATH BIBLE
A KUDO = A Thank you. :-)

Interesting question. I like NOVA, too.

I tested one answer to get a theoretical sense of the question quickly.

Test Answer A. p = 3, yields \(\frac{1}{3}\) + \(\frac{q}{10}\)

q is an integer. There is nothing q can be to make the sum an integer. That is, \(\frac{q}{10}\) cannot be made to equal \(\frac{2}{3}\), e.g., such that sum = 1.

One more quick step to cement the pattern. Plug q = 10 in. Result is \(\frac{1}{3}\) + 1 = \(\frac{4}{3}\) = improper fraction (sum must be integer)

So this result must be avoided: fraction + integer (and vice versa) because result = improper fraction

Because the prompt and answer choices restrict us to controlling the first expression, we must guarantee that it is not a fraction.

The only way to do that is if p is a multiple of 9. (If p = 9 or multiple of 9, q = 10 or multiple of 10.)

Answer C
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If p and q are both positive integers such that p/9 + q/10 is also an   [#permalink] 26 Jul 2017, 10:14
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