Puzzle for March 1, 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 C and D from both sides of eq.5: A + B + C + D – C – D = F – C – C – D which becomes A + B = F – 2×C – D In eq.2, replace A + B with F – 2×C – D: C – E + F = F – 2×C – D Subtract F and C from both sides of the above equation: C – E + F – F – C = F – 2×C – D – F – C which becomes –E = –3×C – D Multiply both sides by (–1): (–1) × (–E) = (–1) × (–3×C – D) which becomes eq.2a) E = 3×C + D
Hint #2
In eq.3, replace E with 3×C + D (from eq.2a): D + 3×C + D = A + C which becomes 3×C + 2×D = A + C Subtract C from both sides of the equation above: 3×C + 2×D – C = A + C – C which becomes eq.3a) 2×C + 2×D = A
Hint #3
Add E to each side of eq.6: F – E + E = B + D + E + E which becomes F = B + D + 2×E In eq.4, substitute B + D + 2×E for F: B + B + D + 2×E = A + D + E which becomes 2×B + D + 2×E = A + D + E Subtract D and E from each side of the above equation: 2×B + D + 2×E – D – E = A + D + E – D – E which becomes eq.4a) 2×B + E = A
Hint #4
Substitute 2×B + E for A (from eq.4a) in eq.3: D + E = 2×B + E + C Subtract E and C from each side of the equation above: D + E – E – C = 2×B + E + C – E – C which becomes D – C = 2×B Divide both sides by 2: (D – C) ÷ 2 = 2×B ÷ 2 which becomes eq.3b) ½×D – ½×C = B
Hint #5
Substitute (3×C + D) for E (from eq.2a), 2×C + 2×D for A (from eq.3a), and ½×D – ½×C for B (from eq.3b) in eq.2: C – (3×C + D) + F = 2×C + 2×D + ½×D – ½×C which is equivalent to C – 3×C – D + F = 1½×C + 2½×D which becomes –2×C – D + F = 1½×C + 2½×D Add 2×C and D to each side of the equation above: –2×C – D + F + 2×C + D = 1½×C + 2½×D + 2×C + D which becomes eq.2b) F = 3½×C + 3½×D
Hint #6
Substitute ½×D – ½×C for B (from eq.3b), 3½×C + 3½×D for F (from eq.2b), 2×C + 2×D for A (from eq.3a), and 3×C + D for E (from eq.2a) in eq.4: ½×D – ½×C + 3½×C + 3½×D = 2×C + 2×D + D + 3×C + D which becomes 3×C + 4×D = 5×C + 4×D Subtract 3×C and 4×D from each side of the above equation: 3×C + 4×D – 3×C – 4×D = 5×C + 4×D – 3×C – 4×D which simplifies to 0 = 2×C which means 0 = C
Hint #7
Substitute 0 for C in eq.3a: 2×0 + 2×D = A which makes 2×D = A
Hint #8
Substitute 0 for C in eq.3b: ½×D – ½×0 = B which means ½×D = B
Hint #9
Substitute 0 for C in eq.2a: E = 3×0 + D which makes E = D
Hint #10
Substitute 0 for C in eq.2b: F = 3½×0 + 3½×D which makes F = 3½×D
Solution
Substitute 2×D for A, ½×D for B, 0 for C, D for E, and 3½×D for F in eq.1: 2×D + ½×D + 0 + D + D + 3½×D = 16 which simplifies to 8×D = 16 Divide both sides of the equation above by 8: 8×D ÷ 8 = 16 ÷ 8 which means D = 2 making A = 2×D = 2 × 2 = 4 B = ½×D = ½ × 2 = 1 E = D = 2 F = 3½×D = 3½ × 2 = 7 and ABCDEF = 410227