Puzzle for March 11, 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
In eq.4, replace A + B with F – A (from eq.3): E + F = F – A + D Subtract F from both sides of the above equation: E + F – F = F – A + D – F which becomes E = –A + D which may be written as eq.4a) E = D – A
Hint #2
In eq.2, replace E with D – A (from eq.4a): B + D – A = A + D In the equation above, subtract D from each side, and add A to each side: B + D – A – D + A = A + D – D + A which makes B = 2×A
Hint #3
In eq.3, substitute 2×A for B: A + 2×A = F – A which becomes 3×A = F – A Add A to both sides of the above equation: 3×A + A = F – A + A which makes 4×A = F
Hint #4
Substitute 4×A for F in eq.6: C – E + 4×A = E – A – C Add E, A, and C to both sides of the above equation: C – E + 4×A + E + A + C = E – A – C + E + A + C which becomes 2×C + 5×A = 2×E Divide both sides by 2: (2×C + 5×A) ÷ 2 = 2×E ÷ 2 which becomes eq.6a) C + 2½×A = E
Hint #5
Substitute C + 2½×A for E (from eq.6a) into eq.4a: C + 2½×A = D – A Add A to both sides of the equation above: C + 2½×A + A = D – A + A which becomes eq.2a) C + 3½×A = D
Hint #6
Substitute C + 3½×A for D (from eq.2a), C + 2½×A for E (from eq.6a), 2×A for B, and 4×A for F in eq.5: C + 3½×A + C + 2½×A = 2×A + C + 4×A which becomes 2×C + 6×A = 6×A + C Subtract 6×A and C from each side of the equation above: 2×C + 6×A – 6×A – C = 6×A + C – 6×A – C which simplifies to C = 0
Hint #7
Substitute 0 for C in eq.2a: 0 + 3½×A = D which makes 3½×A = D
Hint #8
Substitute 0 for C in eq.6a: 0 + 2½×A = E which makes 2½×A = E
Solution
Substitute 2×A for B, 0 for C, 3½×A for D, 2½×A for E, and 4×A for F in eq.1: A + 2×A + 0 + 3½×A + 2½×A + 4×A = 26 which simplifies to 13×A = 26 Divide both sides of the above equation by 13: 13×A ÷ 13 = 26 ÷ 13 which means A = 2 making B = 2×A = 2 × 2 = 4 D = 3½×A = 3½ × 2 = 7 E = 2½×A = 2½ × 2 = 5 F = 4×A = 4 × 2 = 8 and ABCDEF = 240758