In the following cryptarithmetic equation, each letter stands for one unique…

Concept: In a cryptarithmetic (alphametic) addition puzzle, every distinct letter stands for one unique digit from 0-9, and no two letters can share the same digit. Solve it column by column starting from the units place, tracking the carry generated into the next column, and use the bounds those carries place on each letter to pin down its digit.

1. Label the carries c1 (units into tens), c2 (tens into hundreds), and c3 (hundreds into thousands). Since neither NO nor GUN has a thousands digit, the thousands digit of the sum equals c3 exactly, so H = c3. The hundreds column sum is only G + c2, at most 9 + 2 = 11, so c3 can only be 1 -- hence H = 1.

2. Hundreds column: G + c2 = U + 10 times c3 = U + 10 (since c3 = 1). Because G is at most 9 and c2 is at most 2, U = G + c2 - 10 can be at most 1. Since H already takes the digit 1, U must be 0, which forces c2 = 1 and G = 9.

3. Tens column: N + U + N + c1 must produce the tens digit N with carry c2, i.e. N + U + c1 = 10 times c2 = 10. With U = 0, this gives N + c1 = 10, so N = 10 - c1 for c1 in {0, 1, 2}; N = 10 is impossible and N = 9 duplicates G, leaving N = 8 with c1 = 2.

4. Units column: O + N + O = T + 10 times c1, i.e. 2O + 8 = T + 20, so 2O = T + 12. Testing digits not already used (0, 1, 8, 9 are taken): T = 2 gives O = 7, which is valid and distinct from every other letter; no other even T in range works without repeating a digit.

Cross-check: substituting H=1, U=0, N=8, T=2, G=9, O=7 back into the original puzzle: NO = 87, GUN = 908, and 87 + 908 + 87 = 1082, which reads as H U N T = 1, 0, 8, 2 -- matching exactly, with all six letters mapped to six distinct digits.

So HUNT = 1082.