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Find the number of integral solutions of the equation $x\cdot y\cdot z\cdot w=210$

MY TRY :

$210=2\cdot 3\cdot 5\cdot 7$ Considering $x,y,z,w$ as boxes,I get $4^4$ ways since each of the factors can be put in any of the 4 boxes.But that's not the answer.Where am I going wrong?

The answer is $8\cdot 4^4$.

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    $\begingroup$ E.g. $x=-2$, $y=-3$, $z=5$, $w=7$ is also an integral solution, as is $x=1$, $y=1$, $z=2\cdot 3$, $w=5\cdot 7$. $\endgroup$ – Mankind Jan 13 '16 at 13:30
  • $\begingroup$ Oh hey sorry...i was looking at another similar sum while typing and made an error $\endgroup$ – user220382 Jan 13 '16 at 13:31
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There are four boxes available, so by your logic the answer would be $4^4$.

Now, notice that it does not specify that the integers all be positive.

Choose whether $x$ is positive or negative. Choose whether $y$ is positive or negative. Choose whether $z$ is positive or negative.

Depending on these choices, we know that $w$ will either be positive or negative depending on the sign of $xyz$, so there is no choice available here.

Applying multiplication principle, this brings us to a final total of $2^3\cdot 4^4$ number of solutions.

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  • $\begingroup$ Great :-D!!!!!! $\endgroup$ – user220382 Jan 13 '16 at 13:34

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