As it turns out, it is not possible to end up with 105
piles of 1 chip each, using the two operations.
To see this, note that if an odd number n divides evenly
into sizes of all of the piles, then it will still divide
evenly into the sizes after either of the operations has
For the first operation, if an odd n divides evenly into the
sizes of two piles, then it will divide evenly into the sum.
For the second operation, if an odd n divides evenly into
a pile of size 2k, then it must divide evenly into k,
which is the size of each of the
two piles resulting from the split.
Now, let's note that the first operation must be a merge,
since none of the starting piles have an even number of
If we merge the 5 and 49 piles, then we get piles of size
54 and 51, both of which are divisible by 3.
If we merge the 5 and 51 piles, we get sizes 49 and 56,
both divisible by 7.
And if we merge the 49 and 51 piles, we get sizes 5 and
100, which are both divisible by 5.
Hence no matter what subsequent sequence of operations
are performed, the size of each pile will always be
divisible by 3, 5 or 7, and so cannot equal 1.