USA + USSR = PEACE ; P + E + A + C + E = ?

Concept: A cryptarithmetic (alphametic) puzzle assigns a unique digit (0-9) to each distinct letter so that the letters, substituted by their digits, make the addition true, with no number allowed to start with a leading zero. The standard method is to work column by column from the units place, using the number of digits in each word and the carries generated in each column to pin down every letter's digit, testing candidate values against every linked equation until only one consistent assignment survives.

Application:

1. USA is a 3-digit number and USSR is a 4-digit number, so their sum is at most 999 + 9999 = 10998, a number that can only begin with the digit 1. Since the sum is PEACE, this forces P = 1.

2. Aligning the addition by place value: the units column gives A + R, the tens column gives S + S, the hundreds column gives U + S (plus any carry), and the thousands column gives U (plus any carry) — and this thousands column must land on E in PEACE while carrying a 1 into the ten-thousands place (so that P = 1).

3. Because U is a single digit, the thousands-column carry can add at most 1, so U + (carry-in) = 10 + E is possible only if U = 9 and E = 0, with a carry of 1 arriving from the hundreds column into the thousands column.

4. With U = 9, E = 0 and P = 1 fixed, write the three remaining column equations for S (which repeats three times) and A (which repeats twice): the units column needs A + R = 10 (a carry of 1 is the only way to reach E = 0, since A and R are non-zero distinct digits and cannot sum to 0 directly); the tens column needs 2S + 1 = C + 10·c2 for some carry c2 into the hundreds column; and the hundreds column needs 9 + S + c2 = A + 10 (a carry of 1 must leave the hundreds column, to feed the thousands-column carry already fixed above).

5. If c2 = 1, the hundreds equation reduces to A = S, which is impossible since two different letters can never share a digit — so c2 = 0 is forced, giving A = S − 1 and C = 2S + 1.

6. Testing every digit S could take (S cannot be 0, 1 or 9, since those already belong to E, P and U): S = 2 gives A = 1, clashing with P = 1; S = 3 gives A = 2 and C = 7, both distinct from every digit used so far; S = 4 gives C = 9, clashing with U = 9; S = 5 gives C = 11, not a single digit; and S = 6, 7 or 8 all push C = 2S + 1 to 13 or higher, again not a single digit. Only S = 3 survives every check.

7. With S = 3, A = 2 and C = 7 fixed, the units-column equation A + R = 10 gives R = 8 — distinct from every other assigned digit.

8. So USA = 932 and USSR = 9338.

Cross-check: Adding back confirms the puzzle: 932 + 9338 = 10270, and reading PEACE off this sum gives P = 1, E = 0, A = 2, C = 7, E = 0 — exactly the digits derived above.

Therefore P + E + A + C + E = 1 + 0 + 2 + 7 + 0 = 10.