Each letter in the equation below represents a decimal digit. Different letters stand for different digits and the first letter of a multi-digit number cannot be 0.
Find the value of every letter, or prove that no assignment of digits is possible.
Solution:
Write the four–digit numbers in expanded form:
1000A + 100B + 10C + D \; - \; D \; = \; 1000D + 100C + 10B + A.
Simplifying gives
1000A + 100B + 10C = 1000D + 100C + 10B + A,
or
999A + 90B - 90C = 1000D.
Divide both sides by 9:
111A + 10B - 10C = \dfrac{1000D}{9}.
Because the left side is an integer, the right side must be an integer as well, so 1000D must be divisible by 9. The only one–digit values of D that satisfy this are D = 0 or D = 9.
Case 1 D = 0. Then
111A + 10B - 10C = 0 ⇒ 111A = 10(C - B).
The right side is at most 90 in magnitude, so A must be 0. But a leading digit cannot be 0, so this case is impossible.
Case 2 D = 9. Then
111A + 10B - 10C = 9000.
Rearrange:
111A = 9000 - 10(B - C).
The left side is a multiple of 111 = 3·37, so the right side must also be a multiple of 37. Compute 9000 mod 37:
9000 = 37·243 + 9, so 9000 &equiv 9 (mod 37).
Thus
9 - 10(B - C) &equiv 0 &pmod{37} ⇒ 10(B - C) &equiv 9 &pmod{37}.
The modular inverse of 10 mod 37 is 26, giving
B - C &equiv 12 &pmod{37}.
Because B and C are digits, B - C is between –9 and 9, which can never be congruent to 12 modulo 37. Hence this case is also impossible.
Since both possible values of D lead to contradictions, there is no way to assign digits to A, B, C, and D that satisfies the original equation.
Therefore the puzzle has no solution.
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