In the simple subtraction problem below some single digits are replaced by…

Arrange the subtraction as: A 5 C 5 − 1 B 8 7 = 6 7 4 D

1. Units column: 5 − 7 requires a borrow, so 15 − 7 = 8. Therefore D = 8.

2. Tens column: after the borrow the top tens digit is C − 1. We have 10 + (C − 1) − 8 = 4, so C + 1 = 4 and hence C = 3.

3. Hundreds column: because we borrowed for the tens, the top hundreds digit becomes 5 − 1 = 4. Then 10 + 4 − B = 7 (we borrow from the thousands), so 14 − B = 7 and B = 7.

4. Thousands column: after the borrow the top thousands digit is A − 1. Then (A − 1) − 1 = 6 ⇒ A − 2 = 6 ⇒ A = 8.

So the letters are A = 8, B = 7, C = 3, D = 8.

Compute the expression 6A + 4D + 3CD, interpreting 3CD as 3 × C × D:

• 6 × A = 6 × 8 = 48

• 4 × D = 4 × 8 = 32

• 3 × C × D = 3 × 3 × 8 = 72

• Total = 48 + 32 + 72 = 152