The following questions are based on the following multiplication, where each…

In a cryptarithmetic multiplication, every letter stands for one fixed digit throughout the whole diagram. Long multiplication by a two-digit number produces two partial products — one from the multiplier's units digit (unshifted) and one from its tens digit (shifted one place left) — and the digit in each column of the final sum must match the target letter exactly, so working column-by-column with carries pins the letters down.

1. The multiplier BB has the same digit, B, in both its tens and units place. The first (unshifted) partial product is written as the two-digit number “JE” itself, so JE × B = JE — this equality holds only when B = 1.

2. The second partial product (also JE × B, i.e. JE × 1 = JE) is shifted one place left before being added, and is written with three digits, “JEA”. That extra digit A is simply the blank units column created by the shift, so A = 0.

3. Adding the unshifted row “JE” to the shifted row “JE0” gives the four-digit total “BADE”. Matching the hundreds column forces J plus the incoming carry to total 10 (to produce the new leading digit and carry into the thousands place), which is only possible for the single digit J = 9.

4. Matching the tens and units columns of the same addition gives E + J = D + 10 (i.e. E + 9 = D + 10), so E − D = 1 — E and D must be consecutive digits, but the puzzle does not pin down which pair; that ambiguity does not affect the quantity asked.

5. Since B = 1, A = 0, and J = 9 are each uniquely forced, J + A + B = 9 + 0 + 1 = 10 — independent of whichever valid (E, D) pair is chosen.

Direct substitution confirms the whole diagram: with J = 9, E = 3, B = 1, D = 2 (a valid consecutive E−D pair, distinct from J, A, B), JE = 93 and BB = 11, and 93 × 11 = 1023 = B A D E (1, 0, 2, 3) — exactly the target pattern, with the two partial products 93 and 930 adding to 1023.

Hence J + A + B = 10.