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Given that $5x=3y=z$ where $x, y, z$ are integers, each choice given below must be integer except?

A) $z/15 \quad\quad $ B) $z/5 \quad\quad$ C) $z/3 \quad\quad\\ $ D) $z/xy \quad\quad\\$ E) $x/3$


How would I solve this problem any suggestions ?

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The best way to solve this type of problem is to plug some numbers in, and try to get a feel for WHY the various expressions either have to be integers, or don't have to be. It shouldn't take very long to discover that $x$ must be a multiple of 3, $y$ must be a multiple of 5, and that we can write $x=3w$, $y=5w$ and $z=15w$ for some integer $w$.

Since this is clearly a question in a quick-fire multiple choice test, a rigorous proof is unlikely to be needed. In fact, we've already done enough to show that A, B, C and E are not the correct answers. So circle D and move on to the next question. If "none of the above" had been one of the options, then keep going - eventually you'll try $w=2$, and discover that D is indeed correct.

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Why $15$ divides $z$ (meaning that $z/15$ is an integer): If a prime, in this case $3$, divides a product, in this case $5x$, then the prime must divide one (or both) of the terms. Since $3$ does not divide $5$, we conclude that $3$ must divide $x$.

So $x=3t$ for some $t$, and therefore $x=15t$. We conclude that $15$ divides $z$.

The same kind of reasoning (but easier) shows that $z/5$, $z/3$, and $x/3$ are integers. In fact, we have already shown all these things in the process of showing that $15$ divides $z$. So by a process of elimination, the right answer must be D).

But we can see separately that $z/(xy)$ is not necessarily an integer. For example, let $x=6$, $y=10$, and $z=30$. Then your conditions are met, but $xy$, which is $60$, does not divide $z$.

Note that $z/(xy)$ can be an integer, if $x=3$ and $y=5$. These are in fact the only positive cases when under our conditions, $z/(xy)$ is an integer.

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I can see z is divisible by all other options however for $\frac{z}{xy}$ I am assuming $x = 3 \times 5 $ and $y=5 \times 3$ it is also divisible . Am I wrong ? – MistyD Jul 20 '12 at 6:39
If $x=3\times 5$ and $y=5\times 3$, then it is not true that $5x=3y$. I think you meant $x=3$, $y=5$. That certainly is a possibility, and then $z/xy$ is an integer. But there are other possibilities for $x$ and $y$, like the $6$ and $10$ I mentioned. With those, and $z=30$, it will be true that $5x=3y=z$. But for that choice of $x$ and $y$, the number $z/(xy)$ is $30/60$, which is not an integer. So it is not true that $z/(xy)$ must be an integer. (But it can be.) – André Nicolas Jul 20 '12 at 6:50
Thanks that helps – MistyD Jul 20 '12 at 6:57

We have $5x=3y$, or $x \div y = 3\div 5$

Let their greatest common divisor be $k$, where $k$ is constant ;

So $x=3k$, $y=5k$, we have $z=5x=5*3k=15k$; so $z/xy =15k/(3k*5k)=1/k$; and $1/k < 1$, as $k$ is constant $1/k$ is not an integer

follow the same style for other options and you will found the answers are integer,

Thus and (d) $z/xy$ is correct;

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