ABCD
+EFGH
7785
The first two digits of 7785 are 7's.
Since we can't use 9, the largest sum of two digits
is 8+7 which is 15, which ends in 5. So there is
no way to get the two 7's in the beginning of the
7785, by carrying 1 from the preceding column, so
there can be no carrying to the first two columns.
So A+E=7 and B+F=7
The only pairs of digits that add to 7 are these three
1+6, 2+5, and 3+4.
The third digit is 8, which is just 1 more than 7.
That suggests that there was 1 to carry after
adding D+H and getting 15.
The only digits that aren't used in the three pairs
of digits that have sum 7 are the digits 7 and 8.
We notice that they have sum 15.
So we put the 7 and 8 on the right end, then there
is 1 to carry to the third column.
1
ABC7
+EFG8
7785
Now all that's left is to use those three pairs of
digits to fill in the other three pairs of digits.
There are lots of ways to arrange those three pairs
of digits that have sum 7.
We can pick A+E=7 to be 1+6
Then pick B+F=7 to be 2+5
Finally C+G=7 would be 3+4
So one answer is
1
1237
+6548
7785
You can swap these around any way you like,
but there will always be a 7 and 8 on the
right end, and the other three columns
will be the only three possible ways to get 7.
For a couple of examples, you can arrange them
these ways:
1 1
4657 5368
+3128 +2417
7785 7785
or any of many other ways. There are, in fact,
96 different ways to arrange the digits! That
is a combinatorial problem which you will
eventually study if you take enough mathematics.
Yes I know, TMI, :)
Edwin
