Question 263811
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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:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ \ Q\,WW]
*[tex \LARGE \ \ \ \ \ \ \ \ \ \ +WWQ]
*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ \overline{1\,QW\,1}]


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:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ \ \,299]
*[tex \LARGE \ \ \ \ \ \ \ \ \ \ +992]
*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ \,\overline{1291}]


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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