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My son was given a challenge division sheet to be done during a school lesson.

He said he struggled for ages with question 14:

Put brackets into this expression to make it correct.

$10^2 ÷ 10 ÷ 10 ÷ 10 ÷ 10 = 10$

I tried to help but I could not see any way to make this work - it seems you will always end up with an even power of 10 on the left hand side. Can anyone tell us what we are missing?

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    $\begingroup$ Here is a stupid answer: $(10^2 ÷ 10 ÷ 10 ÷ 10 ÷ 1)(0) = (1)(0)$. I agree that the "even power of 10" issue is going to be difficult without some trick. $\endgroup$ – Eric Stucky Oct 7 '16 at 16:07
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    $\begingroup$ Probably a typo in the problem set. $\endgroup$ – John Bentin Oct 7 '16 at 16:08
  • $\begingroup$ Maybe assuming $() \equiv 0$... $\endgroup$ – dxiv Oct 7 '16 at 16:14
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NOTE: This answer assumes cheating, like putting brackets in between digits isn't allowed.

There's a reason why you always get and even power of $10$ on the LHS. Note that by putting brackets on the LHS you actually decide whether you will divide or multiply the product up to that point with $10$. As you have $4$ $10$'s on the LHS you will have the same parity of $10$'s dividing and multiplying $10^2$, as $4$ is an even number. Eventually you will multiply $10^2$ with an even power of $10$. So no matter what, when you multiply/divide an even power with an even power you will get an even power. So eventually the LHS will be an even power of $10$ meaning this question is impossible to solve.

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You can put a single bracket over the equals sign so that it looks a bit like this:

$$10^2÷10÷10÷10÷10 \neq 10$$

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$$0=1(0^2)÷10÷10÷10÷10=1(0)=0$$

This assuming that, as it usually is, multiplication is implicitly assumed; $1(0)=1\times 0$.

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