Puzzle for August 26, 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.
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Hint #1
In eq.2, replace B with A + C (from eq.3): eq.2a) D = A + C + F
Hint #2
In eq.4, replace D with A + C + F (from eq.2a): C + A + C + F = E + F – C which becomes 2×C + A + F = E + F – C In the above equation, subtract F from both sides, and add C to both sides: 2×C + A + F – F + C = E + F – C – F + C which becomes eq.4a) 3×C + A = E
Hint #3
In eq.5, substitute B + F for D (from eq.2): B + E = A + B + F + F which becomes B + E = A + B + 2×F Subtract B from each side of the equation above: B + E – B = A + B + 2×F – B which becomes eq.5a) E = A + 2×F
Hint #4
In eq.5a, substitute 3×C + A for E (from eq.4a): 3×C + A = A + 2×F Subtract A from each side of the above equation: 3×C + A – A = A + 2×F – A which becomes 3×C = 2×F Divide both sides by 2: 3×C ÷ 2 = 2×F ÷ 2 which makes 1½×C = F
Hint #5
eq.6 may be written as: B = (A + E + F) ÷ 3 Multiply both sides of the above equation by 3: 3 × B = 3 × (A + E + F) ÷ 3 which becomes eq.6a) 3×B = A + E + F
Hint #6
In eq.6a, substitute (A + C) for B (from eq.3), and 1½×C for F: 3×(A + C) = A + E + 1½×C which may be written as 3×A + 3×C = A + E + 1½×C Subtract A and 1½×C from both sides of the equation above: 3×A + 3×C – A – 1½×C = A + E + 1½×C – A – 1½×C which becomes eq.6b) 2×A + 1½×C = E
Hint #7
Substitute 2×A + 1½×C for E (from eq.6b) into eq.4a: 3×C + A = 2×A + 1½×C Subtract A and 1½×C from both sides of the equation above: 3×C + A – A – 1½×C = 2×A + 1½×C – A – 1½×C which simplifies to 1½×C = A
Hint #8
Substitute 1½×C for A in eq.4a: 3×C + 1½×C = E which makes 4½×C = E
Hint #9
Substitute 1½×C for A in eq.3: B = 1½×C + C which makes B = 2½×C
Hint #10
Substitute 2½×C for B, and 1½×C for F in eq.2: D = 2½×C + 1½×C which makes D = 4×C
Solution
Substitute 1½×C for A and F, 2½×C for B, 4×C for D, and 4½×C for E in eq.1: 1½×C + 2½×C + C + 4×C + 4½×C + 1½×C = 30 which simplifies to 15×C = 30 Divide both sides of the above equation by 15: 15×C ÷ 15 = 30 ÷ 15 which means C = 2 making A = F = 1½×C = 1½ × 2 = 3 B = 2½×C = 2½ × 2 = 5 D = 4×C = 4 × 2 = 8 E = 4½×C = 4½ × 2 = 9 and ABCDEF = 352893