Thanks EP, guest and Heureka,
Here is another take, it is similar to EPs answer only with no trial and error.
It does not use trial and error.
We know that
\(7x+53=8y\) where x is the number in each original pile and y is the number in the end piles. So obviously x and y are positive integers.
Rearranging I get
\(8y-7x=53\)
One obvious solution is \(x=53\;\; and \;\;y=53\)
so I can say
\(8(53 ) - 7(53 ) = 53\)
\(8(53+7t ) - 7(53 +8t) = 53 \) this is true becasue I have simply added 56t and then taken it away again.
so
\(y=53+7t \;\; and\;\; x=53+8t\)
Now you are also told that at the end you have more than 200 coins
so\(8y>200\\ y>25\)
this means that
\(53+7t>25\\ 7t>25-53\\ 7t>-28\\ t>-4\\ \text{So lets try} \;\; t=-3\\~\\ y=53-21=32\\ x=53-24=29\)
8*32-7*29 = 53 excellent
Initially you had \(7x\) coins. the smallest number this could have been was \(7*29 = 203\) coins