Puzzle for December 23, 2020 ( )
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.6, replace E with A + C (from eq.5): C + A + C = A + D which becomes A + 2×C = A + D Subtract A from both sides of the above equation: A + 2×C – A = A + D – A which makes eq.6a) 2×C = D
Hint #2
In eq.2, replace E with A + C (from eq.5): B + C = A + A + C which becomes B + C = 2×A + C Subtract C from each side of the above equation: B + C – C = 2×A + C – C which makes eq.2a) B = 2×A
Hint #3
In eq.4, replace E with A + C (from eq.5), and D with 2×C: 2×A + A + C = C + 2×C + F which becomes 3×A + C = 3×C + F Subtract C from both sides of the equation above: 3×A + C – C = 3×C + F – C which becomes eq.4a) 3×A = 2×C + F
Hint #4
In eq.4a, substitute (A + F) for C (from eq.3): 3×A = 2×(A + F) + F which is equivalent to 3×A = 2×A + 2×F + F which becomes 3×A = 2×A + 3×F Subtract 2×A from both sides of the above equation: 3×A – 2×A = 2×A + 3×F – 2×A which makes A = 3×F
Hint #5
Substitute (3×F) for A in eq.2a: B = 2×(3×F) which makes B = 6×F
Hint #6
Substitute 3×F for A in eq.3: C = 3×F + F which makes C = 4×F
Hint #7
Substitute (4×F) for C in eq.6a: 2×(4×F) = D which makes 8×F = D
Hint #8
Substitute 3×F for A, and 4×F for C in eq.5: E = 3×F + 4×F which makes E = 7×F
Solution
Substitute 3×F for A, 6×F for B, 4×F for C, 8×F for D, and 7×F for E in eq.1: 3×F + 6×F + 4×F + 8×F + 7×F + F = 29 which simplifies to 29×F = 29 Divide both sides of the above equation by 29: 29×F ÷ 29 = 29 ÷ 29 which means F = 1 making A = 3×F = 3 × 1 = 3 B = 6×F = 6 × 1 = 6 C = 4×F = 4 × 1 = 4 D = 8×F = 8 × 1 = 8 E = 7×F = 7 × 1 = 7 and ABCDEF = 364871