Puzzle for March 25, 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
Multiply each side of eq.4 by both B and F: F ÷ B × B × F = A ÷ F × B × F which becomes eq.4a) F² = A × B Add F to each side of eq.5: (A × B) – F + F = C × F + F which becomes eq.5a) A × B = F × (C + 1)
Hint #2
In eq.5a, replace A × B with F² (from eq.4a): F² = F × (C + 1) Divide both sides of the above equation by F: F² ÷ F = F × (C + 1) ÷ F which makes eq.5b) F = C + 1
Hint #3
In eq.2, replace D with B + C (from eq.1): B + C + E = B + F Subtract B from each side of the above equation: B + C + E – B = B + F – B which becomes eq.2a) C + E = F
Hint #4
In eq.2a, substitute C + 1 for F (from eq.5b): C + E = C + 1 Subtract C from each side of the equation above: C + E – C = C + 1 – C which becomes E = 1
Hint #5
Substitute 1 for E in eq.3: B + 1 = C Subtract 1 from each side of the equation above: B + 1 – 1 = C – 1 which makes eq.3a) B = C – 1
Hint #6
Substitute 1 for E, C – 1 for B (from eq.3a), and C + 1 for F (from eq.5b) in eq.2: D + 1 = C – 1 + C + 1 which becomes D + 1 = 2×C Subtract 1 from both sides of the above equation: D + 1 – 1 = 2×C – 1 which makes eq.2b) D = 2×C – 1
Hint #7
Substitute 2×C – 1 for D (from eq.2b), and C + 1 for F (from eq.5b) in eq.6: B × C = C + 2×C – 1 + C + 1 which becomes B × C = 4×C Divide both sides of the above equation by C: B × C ÷ C = 4×C ÷ C which makes B = 4
Hint #8
Substitute 4 for B in eq.3a: 4 = C – 1 Add 1 to both sides of the above equation: 4 + 1 = C – 1 + 1 which makes 5 = C and makes D = 2×C – 1 = 2×5 – 1 = 10 – 1 = 9 (from eq.2b) F = C + 1 = 5 + 1 = 6 (from eq.5b)
Solution
Substitute 6 for F, and 4 for B in eq.4a: 6² = A × 4 which becomes 36 = A × 4 Divide both sides of the equation above by 4: 36 ÷ 4 = A × 4 ÷ 4 which makes 9 = A and makes ABCDEF = 945916