Find the value of B + A + D if each alphabet represents a unique single digit…

In a cryptarithmetic (alphametic) addition, every distinct letter stands for one unique digit from 0 to 9, and the sum must satisfy ordinary column-by-column addition with carries propagating leftward. Whenever a sum has one more digit than each addend, that extra leading digit can only be the carry generated out of the topmost column, so it must equal 1.

  A B C D
+ E B C B
---------
A F G A G

1. Since AFGAG has five digits while the addends ABCD and EBCB have only four, the leading digit A can only be the carry produced out of the thousands column, so A = 1.

2. The tens digit of the sum is also A = 1, so 2C + (carry from the units column) must equal 1 or 11 (a two-digit tens-column total would itself carry into the hundreds column). Testing 2C + carry = 1 gives C = 0 with a carry of 1 in — but carrying this value through the hundreds and units columns always forces the units-column letter D to come out equal to B, which breaks the rule that every letter is a different digit, so this branch is impossible. That leaves 2C + 1 = 11, giving C = 5, with a carry of 1 arriving from the units column and a carry of 1 passed on to the hundreds column.

3. In the thousands column, 1 + E (plus a carry from the hundreds column) must send a carry of 1 further left, reproducing the leading A = 1. Testing a carry-in of 0 here again forces D past 9, which is impossible, so the hundreds column must carry a 1 in: 1 + E + 1 = F + 10, i.e. F = E − 8. Since A = 1 is already used, taking E = 9 would force F = 1 (repeating A), so the only surviving choice is E = 8 and F = 0.

4. In the hundreds column, 2B + 1(carry) = G + 10 (a carry of 1 must pass on to the thousands column, matching the previous step), so G = 2B − 9. In the units column, D + B = G + 10 (a carry of 1 must leave for the tens column, matching the earlier step), so D = B + 1. Checking every remaining candidate for B (5 through 9) against the digits already fixed (A = 1, C = 5, E = 8, F = 0): B = 5 repeats C; B = 7 gives G = 5, repeating C; B = 8 repeats E; B = 9 gives G = 9, repeating B itself. Only B = 6 survives every check, giving G = 2(6) − 9 = 3 and D = 6 + 1 = 7.

5. This yields the complete, self-consistent assignment: A = 1, B = 6, C = 5, D = 7, E = 8, F = 0, G = 3 — seven letters, seven distinct digits, exactly as the puzzle requires.

Cross-check: ABCD = 1657 and EBCB = 8656; adding them gives 1657 + 8656 = 10313, and AFGAG with these digits reads 1 0 3 1 3 = 10313 — an exact match, confirming the assignment against the original puzzle rather than the column equations alone.

So B + A + D = 6 + 1 + 7 = 14.