Puzzle for February 6, 2021 ( )
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.3, replace D + F with C + E (from eq.5): B + E = C + C + E which becomes B + E = 2×C + E Subtract E from both sides of the equation above: B + E – E = 2×C + E – E which makes B = 2×C
Hint #2
In eq.4, replace B with 2×C: D – C = 2×C – D Add C and D to each side of the equation above: D – C + C + D = 2×C – D + C + D which becomes 2×D = 3×C Divide both sides by 2: 2×D ÷ 2 = 3×C ÷ 2 which makes D = 1½×C
Hint #3
In eq.2, substitute 2×C for B, and 1½×C for D: A + 2×C = 1½×C + E Subtract 1½×C from both sides of the above equation: A + 2×C – 1½×C = 1½×C + E – 1½×C which becomes eq.2a) A + ½×C = E
Hint #4
Substitute A + ½×C for E (from eq.2a), and 1½×C for D in eq.5: C + A + ½×C = 1½×C + F which becomes A + 1½×C = 1½×C + F Subtract 1½×C from both sides of the equation above: A + 1½×C – 1½×C = 1½×C + F – 1½×C which makes A = F
Hint #5
Substitute 2×C for B, 1½×C for D, and (A + ½×C) for E (from eq.2a) in eq.6: (A × 2×C) – 1½×C = (1½×C × (A + ½×C)) – C which is equivalent to (2 × (A×C)) – 1½×C = (1½×C × A) + (1½×C × ½×C) – C which may be written as 2×(A×C) – 1½×C = 1½×(A×C) + ¾×C² – C In the above equation, add 1½×C to both sides, and subtract 1½×(A×C) from both sides: 2×(A×C) – 1½×C + 1½×C – 1½×(A×C) = 1½×(A×C) + ¾×C² – C + 1½×C – 1½×(A×C) which simplifies to eq.6a) ½×(A×C) = ¾×C² + ½×C
Hint #6
Divide both sides of eq.6a by C: ½×(A×C) ÷ C = (¾×C² + ½×C) ÷ C which becomes ½×A = ¾×C + ½ Multiply both sides of the above equation by 2: (½×A) × 2 = (¾×C + ½) × 2 which makes A = 1½×C + 1 and also makes eq.6b) F = A = 1½×C + 1
Hint #7
Substitute 1½×C + 1 for A (from eq.6b) in eq.2a: 1½×C + 1 + ½×C = E which makes eq.2b) 2×C + 1 = E
Solution
Substitute 1½×C + 1 for A and F (from eq.6b), 2×C for B, 1½×C for D, and 2×C + 1 for E (from eq.2b) in eq.1: 1½×C + 1 + 2×C + C + 1½×C + 2×C + 1 + 1½×C + 1 = 41 which simplifies to 9½×C + 3 = 41 Subtract 3 from both sides of the above equation: 9½×C + 3 – 3 = 41 – 3 which makes 9½×C = 38 Divide both sides by 9½: 9½×C ÷ 9½ = 38 ÷ 9½ which means C = 4 making A = F = 1½×C + 1 = 1½×4 + 1 = 6 + 1 = 7 (from eq.6b) B = 2×C = 2 × 4 = 8 D = 1½×C = 1½ × 4 = 6 E = 2×C + 1 = 2×4 + 1 = 8 + 1 = 9 (from eq.2b) and ABCDEF = 784697