If p, q, x, and y are positive integers, is q^y/p^x an integer?

St 1 says q/p= Integer, I so we have q=p*I Subsituting in given eqn we get that (P*I)^y/p^x...clearly we need to know y and x..consider x=100 and p=2,q=8 and y=2...We will get an integer but if x=1,y=2 then an integer...not sufficient.. A and D ruled out

2. says \(y \ge x\) but we don't know whether p is factor of q or not.. B is out

Combining, both are conditions are met so Ans is C and we will get an Integer

Official Solution:


The question asks if we can determine whether the expression \(\frac{q^y}{p^x}\) yields an integer. This is the same as asking if \(q^y\) is divisible by \(p^x\) or, equivalently, if \(p^x\) is a factor of \(q^y\).

Statement 1 tells us that \(q\) is evenly divisible by \(p\), or, in other words, that \(p\) is a factor of \(q\). Let us plug in numbers that satisfy this condition and see if we can answer the question in the prompt. We will choose \(p = 2\) and \(q = 4\). However, since we are given no information on \(x\) or \(y\), we can plug in any integer values for \(x\) and \(y\). If \(x = 3\) and \(y = 1\), then \(\frac{q^y}{p^x} = \frac{4^1}{2^3} = \frac{4}{8} = \frac{1}{2}\), which is not an integer. On the other hand, if \(y = 3\) and \(x = 1\), \(\frac{q^y}{p^x} = \frac{4^3}{2^1} = \frac{64}{2} = 32\), which is an integer. Statement 1 alone is NOT sufficient to answer the question. Eliminate answer choices A and D. The correct answer choice is B, C, or E.

Statement 2 tells us that \(y \ge x\). However, we know nothing about the values of \(p\) and \(q\), which can be any integers. Whether the expression yields an integer depends on the relationship between \(p\) and \(q\). Statement 2 alone is also NOT sufficient to answer the question. Eliminate answer choice B. The correct answer choice is either C or E.

Taking both statements together: Statement 1 tells us that \(p\) is a factor of \(q\), which implies that \(p \le q\), and statement 2 tells us that \(x \le y\). This means that the term \(q^y\) will always be larger than \(p^x\), and since \(p\) is a factor of \(q\), \(\frac{q^y}{p^x}\) will always yield an integer. For example, if \(p = 3\), \(q = 6\), \(x = 1\), and \(y = 2\), then \(\frac{q^y}{p^x} = \frac{6^2}{3^1} = \frac{36}{3} = 12\). Both statements together are sufficient.


Answer: C.

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