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?
On
$$0=1(0^2)÷10÷10÷10÷10=1(0)=0$$
This assuming that, as it usually is, multiplication is implicitly assumed; $1(0)=1\times 0$.
On
You can put a single bracket over the equals sign so that it looks a bit like this:
$$10^2÷10÷10÷10÷10 \neq 10$$
On
You just calculate the whole thing, but instead of multiplying or dividing by ten, you count the powers of ten and you add or subtract them, so you get:
$$ 2 ± 1 ± 1 ± 1 ± 1 = 1$$
Now check what happens with the parity of the result, when you are adding or subtracting one:
$2$ : parity : even
$2± 1$ : parity : odd
$2± 1 ± 1$ : parity : even
$2± 1 ± 1 ± 1$ : parity : even
$2± 1 ± 1 ± 1 ± 1$ : parity : even
And as an $1$ (the final result) is not even, you can see that it's impossible to get this result.
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.