Look at the sum of the 100s digits. The largest Q and W can be is 8 and 9 in some order. Depending on whether or not there was a carry from the previous column, the most Q + W + a possible carry can be is 18, and since the sum is YQ, Y must be 1.
So, now we have:
Q plus W cannot equal 1 because that would mean that one of them had to be zero and the problem says the letters are all non-zero and the other would have to be 1 which is already used by Y and the problem says they are all different.
Therefore Q plus W must be 11.
Q = 9 if and only if W = 2
Q = 8 if and only if W = 3
and so on until we get to
Q = 2 if and only if W = 9
But W + W + 1 must either equal W or 1W so the only possibility is W = 9, so that W + W + 1 = 9 + 9 + 1 = 18. Hence Q = 2.
Therefore:
John