so Since s = 3 --> 3o Since m can't be 3, --> 3o
+many +many m must be 2 and +2any
-------- -------- there must be a --------
sums 3um3 carry. 3u23
Now since o + y = 3, either o = 2 and y =1 or o = 1 and y = 2 (both impossible because n = 2) --> o = 6 and y = 7 or o = 7 and y = 6 (so their sum ends with a 3)
Try o = 6 and y = 7 --> 36 3 + n + 1 (a carry) has a sum that ends with 2;
2an7 the sum must be 12 and n must be 8
--------
3u23
--> 36 --> Now, a + 1 (a carry) must add to u; but this sum has to be at least 10, so there
+ 2a87 is a carry for the next columns to the left: therefore, the sum is exactly 10,
------- so u = 0. This forces a to be 9.
3u23
36
+ 2987
--------
3023 But, this is the answer when o = 6 and y = 7; could y = 6 and o = 7?