Puzzle for December 8, 2022 ( )
Scratchpad
Find the 6-digit number ABCDEF by solving the following equations:
A, B, C, D, E, and F each represent a one-digit non-negative integer.
Scratchpad
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Hint #1
Add B to both sides of eq.5: A - C + D + B = C + F - B + B which becomes A - C + D + B = C + F which may be re-written as eq.5a) A + B - C + D = C + F
Hint #2
In eq.5a, replace A + B with C + E (from eq.3): C + E - C + D = C + F which becomes E + D = C + F which may be re-written as eq.5b) D + E = C + F
Hint #3
In eq.4, replace D + E with C + F (from eq.5b): C + F = B + C + F Subtract C and F from each side of the above equation: C + F - C - F = B + C + F - C - F which simplifies to 0 = B
Hint #4
In eq.2, substitute 0 for B: 0 + D = A - 0 which makes D = A
Hint #5
Substitute D for A, and 0 for B in eq.3: C + E = D + 0 which becomes eq.3a) C + E = D
Hint #6
Substitute C + E for D (from eq.3a), and 0 for B in eq.4: C + E + E = 0 + C + F which becomes C + 2×E = C + F Subtract C from each side of the equation above: C + 2×E - C = C + F - C which makes 2×E = F
Hint #7
eq.6 may be written as: E + F = (A + B + D) ÷ 3 Multiply both sides of the above equation by 3: 3 × (E + F) = 3 × (A + B + D) ÷ 3 which becomes eq.6a) 3×E + 3×F = A + B + D
Hint #8
Substitute (2×E) for F, 0 for B, and A for D in eq.6a: 3×E + 3×(2×E) = A + 0 + A which becomes 3×E + 6×E = 2×A which makes 9×E = 2×A Divide both sides of the above equation by 2: 9×E ÷ 2 = 2×A ÷ 2 which makes 4½×E = A and also makes 4½×E = A = D
Hint #9
Substitute 4½×E for D in eq.3a: C + E = 4½×E Subtract E from both sides of the equation above: C + E - E = 4½×E - E which makes C = 3½×E
Solution
Substitute 4½×E for A and D, 0 for B, 3½×E for C, and 2×E for F in eq.1: 4½×E + 0 + 3½×E + 4½×E + E + 2×E = 31 which simplifies to 15½×E = 31 Divide both sides of the above equation by 15½: 15½×E ÷ 15½ = 31 ÷ 15½ which means E = 2 making A = D = 4½×E = 4½ × 2 = 9 C = 3½×E = 3½ × 2 = 7 F = 2×E = 2 × 2 = 4 and ABCDEF = 907924