Puzzle for December 1, 2019 ( )
Scratchpad
Find the 6-digit number ABCDEF by solving the following equations:
A, B, C, D, E, and F each represent a one-digit positive integer.
Scratchpad
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Hint #1
In eq.2, replace E + F with D – C (from eq.5): B + C + D – C = A + D which becomes B + D = A + D Subtract D from both sides of the above equation: B + D – D = A + D – D which makes B = A
Hint #2
In eq.3, replace A with B: C – F = B – B which becomes C – F = 0 Add F to each side of the equation above: C – F + F = 0 + F which means C = F
Hint #3
In eq.5, substitute C for F: E + C = D – C Add C to both sides of the above equation: E + C + C = D – C + C which becomes eq.5a) 2×C + E = D Subtract E from each side of eq.5a: 2×C + E – E = D – E which becomes eq.5b) 2×C = D – E
Hint #4
Add D, E, and F to both sides of eq.4: D – E – F + D + E + F = B – D + D + E + F which simplifies to 2×D = B + E + F Substitute (2×C + E) for D (from eq.5a), and C for F in the above equation: 2×(2×C + E) = B + E + C which becomes 4×C + 2×E = B + E + C Subtract C and E from each side: 4×C + 2×E – C – E = B + E + C – C – E which becomes 3×C + E = B and also makes eq.4a) A = B = 3×C + E
Hint #5
Substitute 3×C + E for both A and B (from eq.4a) in eq.6: 3×C + E + 3×C + E + D + E = D × E which becomes 6×C + D + 3×E = D × E which may be written as eq.6a) 3×(2×C) + D + 3×E = D × E
Hint #6
Substitute D – E for 2×C (from eq.5b) in eq.6a: 3×(D – E) + D + 3×E = D × E which becomes 3×D – 3×E + D + 3×E = D × E which makes 4×D = D × E Divide both sides of the above equation by D: 4×D ÷ D = D × E ÷ D which means 4 = E
Hint #7
Substitute 4 for E in eq.4a: eq.4b) B = A = 3×C + 4
Hint #8
Substitute 4 for E in eq.5a: eq.5b) 2×C + 4 = D
Solution
Substitute 3×C + 4 for A and B (from eq.4b), 2×C + 4 for D (from eq.5b), 4 for E, and C for F in eq.1: 3×C + 4 + 3×C + 4 + C + 2×C + 4 + 4 + C = 26 which simplifies to 10×C + 16 = 26 Subtract 16 from both sides of the equation above: 10×C + 16 – 16 = 26 – 16 which makes 10×C = 10 Divide both sides by 10: 10×C ÷ 10 = 10 ÷ 10 which means C = 1 making B = A = 3×C + 4 = 3×1 + 4 = 3 + 4 = 7 (from eq.4b) D = 2×C + 4 = 2×1 + 4 = 2 + 4 = 6 (from eq.5b) F = C = 1 and ABCDEF = 771641