geno3141

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 #1
avatar+23252 
+5

       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?

13 oct 2014