Puzzle for May 30, 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
Subtract B and D from each side of eq.3: A + D – B – D = B + C – B – D which becomes eq.3a) A – B = C – D Subtract B, C, and E from both sides of eq.5: B + D – B – C – E = A + C + E – B – C – E which becomes eq.5a) D – C – E = A – B
Hint #2
In eq.5a, replace A – B with C – D (from eq.3a): D – C – E = C – D Add C, D, and E to each side of the equation above: D – C – E + C + D + E = C – D + C + D + E which becomes eq.5b) 2×D = 2×C + E
Hint #3
In eq.2, replace E with D + F (from eq.6), and B with E + F (from eq.4): D + D + F = A + E + F which becomes 2×D + F = A + E + F Subtract F from both sides of the equation above: 2×D + F – F = A + E + F – F which becomes eq.2a) 2×D = A + E
Hint #4
Substitute A + E for 2×D (from eq.2a) in eq.5b: A + E = 2×C + E Subtract E from each side of the above equation: A + E – E = 2×C + E – E which makes A = 2×C
Hint #5
In eq.5, substitute E + F for B (from eq.4): E + F + D = A + C + E Subtract E from each side of the above equation: E + F + D – E = A + C + E – E which becomes F + D = A + C which may be written as eq.5c) D + F = A + C
Hint #6
Substitute E for D + F (from eq.6), and 2×C for A in eq.5c: E = 2×C + C which makes E = 3×C
Hint #7
Substitute 2×C for A, and 3×C for E in eq.5: B + D = 2×C + C + 3×C which makes eq.5d) B + D = 6×C
Hint #8
Substitute 2×C for A in eq.3: 2×C + D = B + C Subtract C from each side of the above equation: 2×C + D – C = B + C – C which becomes eq.3b) C + D = B
Hint #9
Substitute C + D for B (from eq.3b) in eq.5d: C + D + D = 6×C which becomes C + 2×D = 6×C Subtract C from each side of the above equation: C + 2×D – C = 6×C – C which makes 2×D = 5×C Divide both sides by 2: 2×D ÷ 2 = 5×C ÷ 2 which makes D = 2½×C
Hint #10
Substitute 2½×C for D in eq.5d: B + 2½×C = 6×C Subtract 2½×C from both sides of the above equation: B + 2½×C – 2½×C = 6×C – 2½×C which makes B = 3½×C
Hint #11
Substitute 3×C for E, and 2½×C for D in eq.6: 3×C = 2½×C + F Subtract 2½×C from both sides of the equation above: 3×C – 2½×C = 2½×C + F – 2½×C which makes ½×C = F
Solution
Substitute 2×C for A, 3½×C for B, 2½×C for D, 3×C for E, and ½×C for F in eq.1: 2×C + 3½×C + C + 2½×C + 3×C + ½×C = 25 which simplifies to 12½×C = 25 Divide both sides of the above equation by 12½: 12½×C ÷ 12½ = 25 ÷ 12½ which means C = 2 making A = 2×C = 2 × 2 = 4 B = 3½×C = 3½ × 2 = 7 D = 2½×C = 2½ × 2 = 5 E = 3×C = 3 × 2 = 6 F = ½×C = ½ × 2 = 1 and ABCDEF = 472561