If (HE)H = SHE, where the letters H, S and E each represent a single digit…

Concept: In a cryptarithmetic equation, every letter stands for exactly one fixed digit (0-9) everywhere it appears, and a letter used as a digit in the number SHE must match the same letter used elsewhere in the equation. When a two-digit number is raised to a power to reproduce a three-digit number sharing its own digits, the place-value range of the three-digit result (100 to 999) sharply limits which power is even possible, so the first move is to bound the exponent before testing any base.

Application:

1. SHE is a 3-digit number, so 100 ≤ (HE)H ≤ 999.

2. If H = 1, (HE)1 = HE, which is only 2 digits — too small. If H ≥ 3, even the smallest base with leading digit H already overshoots 999 when raised to that power (e.g. 303 = 27000). So H = 2 is the only exponent that can keep the result in the 3-digit range.

3. With H = 2, the base HE = 20 + E for some digit E, and (20 + E)2 must be a 3-digit number whose tens digit equals H = 2 and whose units digit equals E.

4. Testing E = 0 through 9, only E = 5 gives a base of 25, and 252 = 625.

5. Checking place values in 625: hundreds digit = 6, tens digit = 2 (matches H), units digit = 5 (matches E) — every position is consistent, so S = 6, H = 2, E = 5.

Cross-check: running the same tens-digit-must-equal-H and units-digit-must-equal-E test across every other E from 0 to 9 (bases 20 through 29 excluding 25) fails at either the tens or the units place — 25² = 625 is the only base in range that satisfies both conditions simultaneously, so the digit assignment is unique.

Result: S + H + E = 6 + 2 + 5 = 13.