Answer Of Alphamatic Problem
Here is question!
First of all let's write down the equation once again.
ABCABA
+
BBDCAA
+
ABEABB
+
ABDBAA
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AAFGBDH
We will refer to places in number from left as a first, second, third...sixth instead of tenth, hundredth, thousandth etc.
First we need to find if the 5 digits of first number itself i.e. ABCAB are carries forwarded from previous place.
From the addition of variable from first place, we get,
3A + B = 10A + A ........(1)
B = 8A ........(2)
Only numbers satisfying above are A = 1 & B = 8 , but at previous place we have addition of 4 B's. If B = 8, then addition at second place would be 32 with F = 2 & carry 3 which is not equal to A = 1. So A can't be a carry. So we need to modify (2) above as
B + x = 8A .........(2)....Where 'x' is carry forwarded from second place.
If B = 1 or 2 then x = 0 as at second place we would have F = 4 or 8. In that case, A would be fractional. Some other possible combinations for B, A & x are,
B = 9, x = 3, 8A = 12,
B = 8, x = 3, 8A = 10,
B = 7, x = 2, 8A = 9,
B = 6, x = 2, 8A = 8,
This is the only combination that can make A a whole number. So A = 1, B = 6.
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From sixth place, we have,
H = 3A + B = 9
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From fifth place, we have,
D = 2A + 2B = 14
But it has to be single digit i.e. D = 4 with 1 carry forwarded to next.
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From fourth place,
B = 2A + B + C + 1 .....1 is carry from last place.
6 = 9 + C
Now C can't be negative hence C + 9 has to be 2 digit number with 6 at last digit.Since addition of 2 single digit numbers never exceeds 18, C + 9 has to be 16.
16 = 9 + C
gives, C = 7 & carry 1 forwarded to third place.
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