TOM + NAG = GOAT, find the value of G+O+A+T.

Concept

In a cryptarithmetic addition, every letter stands for one fixed digit (0-9), no two letters share a digit, and a leading letter of any number cannot be 0. Solve it column by column (units, tens, hundreds, ...) using the carries the addition produces. If, after applying every column constraint, one or more letters still have more than one digit that fits, the puzzle admits multiple valid digit assignments - and a quantity built from those letters is genuinely Cannot be determined only if it actually takes different values across those valid assignments, so each candidate assignment must be checked against the target expression, never assumed to vary.

Application

TOM + NAG = GOAT lines up column by column as: units (M + G), tens (O + A), hundreds (T + N), and a new thousands place that can only come from a final carry.

1. A 3-digit number plus a 3-digit number is at most 999 + 999 = 1998, so the only way the sum becomes a 4-digit number is a carry of 1 into the thousands place - that carry IS the leading digit, so G = 1.

2. The tens column reads O + A (+ any carry from units) and must still end in A. That only works if O plus the incoming carry equals 0 or 10. Testing O = 9 forces M = T + 9, which is impossible for single digits without reusing a digit already taken or making a leading digit 0 - so the only consistent case is O = 0 with no carry out of the units column.

3. With O = 0, the hundreds column (T + N, with a carry of 1 going out to make G = 1) gives T + N = 10.

4. The units column (M + G, with no carry out) gives M = T - 1, i.e. M + 1 = T.

5. Scanning every digit pair with T + N = 10 and M = T - 1, while keeping G = 1, O = 0, T, N, M all distinct, shows more than one combination works - for example T = 6, N = 4, M = 5, or equally T = 3, N = 7, M = 2, or T = 4, N = 6, M = 3.

6. In every one of these valid combinations, A is simply whatever digit is left over and unused - nothing in the addition pins A, or even T itself, to one value.

Cross-check

Plugging T = 6, N = 4, M = 5, O = 0, G = 1 back in: TOM = 605, and NAG = 4A1 for any leftover A. Taking A = 2 gives 605 + 421 = 1026 = GOAT with G=1, O=0, A=2, T=6 - a valid fit. Taking A = 3 instead gives 605 + 431 = 1036, again a valid fit with a different A (and a different G+O+A+T total). Since both assignments satisfy TOM + NAG = GOAT equally well, no single value of A - or of the sum G + O + A + T - is forced by the puzzle.

Result

Because both T and A can each take more than one valid digit while every column constraint stays satisfied, G + O + A + T does not resolve to one fixed number. The answer is Cannot be determined.