Puzzle for January 10, 2021 ( )
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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.
* EF is a 2-digit number (not E×F).
** DEF is a 3-digit number (not D×E×F). BC is a 2-digit number (not B×C).
Our thanks to Judah S (age 14) for contributing this puzzle. Thanks, Judah!
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Hint #1
Multiply both sides of eq.2 by (C × A): (B ÷ C) × (C × A) = (E ÷ A) × (C × A) which becomes B × A = E × C which is the same as A × B = C × E In eq.1, replace A × B with C × E: C × E = C × F Divide both sides of the above equation by C: (C × E) ÷ C = (C × F) ÷ C which makes E = F
Hint #2
In eq.6, substitute (BC × F) for DEF (from eq.4): B × BC = (BC × F) – BC which may be written as B × BC = BC × (F – 1) Divide both sides by BC: (B × BC) ÷ BC = (BC × (F – 1)) ÷ BC which becomes B = F – 1 Add 1 to both sides of the above equation: B + 1 = F – 1 + 1 which makes B + 1 = F and also makes eq.6a) E = F = B + 1
Hint #3
In eq.1, substitute (B + 1) for F (from eq.6a): A × B = C × (B + 1) Divide both sides of the above equation by B: (A × B) ÷ B = (C × (B + 1)) ÷ B which becomes eq.1a) A = (C × (B + 1)) ÷ B
Hint #4
eq.3 may be written as: 10×E + F = A × C – E In the above equation, substitute (B + 1) for both E and F (from eq.6a): 10×(B + 1) + (B + 1) = A × C – (B + 1) which becomes 10×B + 10 + B + 1 = A × C – B – 1 which becomes 11×B + 11 = A × C – B – 1 Add B and 1 to both sides: 11×B + 11 + B + 1 = A × C – B – 1 + B + 1 which becomes eq.3a) 12×B + 12 = A × C
Hint #5
eq.5 may be written as: A × C × B = DEF Substitute (12×B + 12) for A × C (from eq.3a), and BC × F for DEF (from eq.4) in the equation above: (12×B + 12) × B = BC × F which becomes 12×B² + 12×B = BC × F which may be written as eq.5a) 12×B×(B + 1) = (10×B + C) × F
Hint #6
Substitute (B + 1) for F (from eq.6a) in eq.5a: 12×B×(B + 1) = (10×B + C) × (B + 1) Divide both sides of the above equation by (B + 1): (12×B×(B + 1)) ÷ (B + 1) = ((10×B + C) × (B + 1)) ÷ (B + 1) which becomes 12×B = 10×B + C Subtract 10×B from both sides of the above equation: 12×B – 10×B = 10×B + C – 10×B which makes 2×B = C
Hint #7
Substitute 2×B for C in eq.3a: 12×B + 12 = A × 2×B Divide both sides of the equation above by 2×B: (12×B + 12) ÷ 2×B = (A × 2×B) ÷ 2×B which becomes eq.3b) (6×B + 6) ÷ B = A
Hint #8
Substitute (6×B + 6) ÷ B for A (from eq.3b), and 2×B for C in eq.1a: (6×B + 6) ÷ B = (2×B × (B + 1)) ÷ B Multiply both sides of the equation above by B: ((6×B + 6) ÷ B) × B = ((2×B × (B + 1)) ÷ B) × B which becomes eq.1b) 6×B + 6 = 2×B × (B + 1)
Hint #9
eq.1b may be re-written as: 6×(B + 1) = 2×B × (B + 1) Divide both sides of the above equation by (B + 1): (6×(B + 1)) ÷ (B + 1) = (2×B × (B + 1)) ÷ (B + 1) which means 6 = 2×B Divide both sides by 2: 6 ÷ 2 = 2×B ÷ 2 which makes 3 = B and makes C = 2×B = 2 × 3 = 6
Hint #10
Substitute 3 for B in eq.3b: (6×3 + 6) ÷ 3 = A which becomes (18 + 6) ÷ 3 = A which makes 24 ÷ 3 = A which means 8 = A
Hint #11
Substitute 3 for B in eq.6a: E = F = 3 + 1 which makes E = F = 4
Solution
eq.5 may be written as: A × B × C = 100×D + 10×E + F In the above equation, substitute 8 for A, 3 for B, 6 for C, and 4 for E and F: 8 × 3 × 6 = 100×D + 10×4 + 4 which becomes 144 = 100×D + 40 + 4 which becomes 144 = 100×D + 44 Subtract 44 from both sides: 144 – 44 = 100×D + 44 – 44 which makes 100 = 100×D Divide both sides by 100: 100 ÷ 100 = 100×D ÷ 100 which makes 1 = D and makes ABCDEF = 836144