Puzzle for August 15, 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 non-negative integer.
* EF is a 2-digit number (not E×F).
Our thanks to Tom H for sending us this challenging puzzle. Thank you, Tom!
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Hint #1
Add F to both sides of eq.3: (C × E) – F + F = D + F which becomes (C × E) = D + F In eq.1, replace D + F with (C × E): A = (C + (C × E)) ÷ C which is equivalent to A = (C ÷ C) + ((C × E) ÷ C) which becomes A = (1 + (1 × E)) which is the same as eq.1a) A = 1 + E
Hint #2
In eq.2, replace A with 1 + E (from eq.1a): B = 1 + E + E which makes eq.2a) B = 1 + 2×E
Hint #3
eq.6 may be re-written as: F = (F × E) – (A × E) In the above equation, add (A × E) to both sides, and subtract F from both sides: F + (A × E) – F = (F × E) – (A × E) + (A × E) – F which becomes A × E = (F × E) – F which is equivalent to eq.6a) A × E = F × (E – 1)
Hint #4
In eq.6a, substitute (1 + E) for A (from eq.1a): (1 + E) × E = F × (E – 1) which becomes E + E² = F × (E – 1) Divide both sides of the above equation by (E – 1): (E + E²) ÷ (E – 1) = F × (E – 1) ÷ (E – 1) which becomes eq.6b) (E + E²) ÷ (E – 1) = F (assumes E ≠ 1)
Hint #5
Confirm: E ≠ 1 ... If E = 1, then substituting 1 for E in eq.1a would yield: A = 1 + 1 which would make A = 2 Substituting 2 for A, and 1 for E in eq.6 would yield: F = (F – 2) × 1 which would mean F = F – 2 However: F ≠ F – 2 So, therefore: E ≠ 1
Hint #6
eq.5 may be written as: 10×E + F = A × D In the above equation, substitute ((E + E²) ÷ (E – 1)) for F (from eq.6b), and (1 + E) for A (from eq.1a): 10×E + ((E + E²) ÷ (E – 1)) = (1 + E) × D which is equivalent to eq.5a) 10×E + ((E × (1 + E)) ÷ (E – 1)) = (1 + E) × D
Hint #7
Divide both sides of eq.5a by (1 + E): (10×E + ((E × (1 + E)) ÷ (E – 1))) ÷ (1 + E) = ((1 + E) × D) ÷ (1 + E) which becomes (10×E ÷ (1 + E)) + ((E × (1 + E)) ÷ ((E – 1) × (1 + E))) = D which becomes eq.5b) (10×E ÷ (1 + E)) + (E ÷ (E – 1)) = D
Hint #8
Substitute ((E + E²) ÷ (E – 1)) for F (from eq.6b) in eq.4: D ÷ E = ((E + E²) ÷ (E – 1)) – E In the above equation, multiply the final E by (E – 1) ÷ (E – 1). This will give the same common denominator to both terms in the right side. D ÷ E = ((E + E²) ÷ (E – 1)) – (E × (E – 1) ÷ (E – 1)) which becomes D ÷ E = ((E + E²) ÷ (E – 1)) – ((E² – E) ÷ (E – 1)) which becomes D ÷ E = ((E + E²) – (E² – E)) ÷ (E – 1) which becomes D ÷ E = (E + E² – E² + E) ÷ (E – 1) which becomes D ÷ E = 2×E ÷ (E – 1) Multiply both sides by E: (D ÷ E) × E = (2×E ÷ (E – 1)) × E which becomes eq.4a) D = 2×E² ÷ (E – 1)
Hint #9
Substitute (2×E² ÷ (E – 1)) for D (from eq.4a) in eq.5b: (10×E ÷ (1 + E)) + (E ÷ (E – 1)) = (2×E² ÷ (E – 1)) Subtract (E ÷ (E – 1)) from each side of the equation above: (10×E ÷ (1 + E)) + (E ÷ (E – 1)) – (E ÷ (E – 1)) = (2×E² ÷ (E – 1)) – (E ÷ (E – 1)) which becomes eq.5c) 10×E ÷ (1 + E) = (2×E² – E) ÷ (E – 1)
Hint #10
Since the fractions on both sides of eq.5c are equal, their cross-product will also be equal. Calculate the cross-product of these fractions by cross-multiplying their numerators with their denominators: 10×E × (E – 1) = (2×E² – E) × (1 + E) which becomes 10×E² – 10×E = 2×E² – E + 2×E³ – E² which becomes 10×E² – 10×E = E² – E + 2×E³ Since E ≠ 0 (from eq.4), divide both sides of the equation above by E: (10×E² – 10×E) ÷ E = (E² – E + 2×E³) ÷ E which becomes 10×E – 10 = E – 1 + 2×E² In the above equation, subtract 10×E from both sides, and add 10 to both sides: 10×E – 10 – 10×E + 10 = E – 1 + 2×E² – 10×E + 10 which becomes eq.5d) 0 = 2×E² – 9×E + 9
Hint #11
eq.5d is a quadratic equation in standard form. Using the quadratic equation solution formula to solve for E in eq.5d yields: E = { (–1)×(–9) ± sq.rt.[(–9)² – (4 × 2 × 9)] } ÷ (2 × 2) which becomes E = {9 ± sq.rt.(81 – 72)} ÷ 4 which becomes E = (9 ± sq.rt.(9)) ÷ 4 which becomes E = (9 ± 3) ÷ 4 In the above equation, either E = (9 + 3) ÷ 4 = 12 ÷ 4 = 3 or E = (9 – 3) ÷ 4 = 6 ÷ 4 = 1½ Since E must be an integer, then E ≠ 1½ and therefore makes E = 3
Hint #12
Substituting 3 for E in eq.1a, eq.2a, eq.4a, and eq.6b makes: A = 1 + E = 1 + 3 = 4 (from eq.1a) B = 1 + 2×3 = 1 + 2×E = 1 + 6 = 7 (from eq.2a) D = 2×E² ÷ (E – 1) = 2×3² ÷ (3 – 1) = 2×9 ÷ 2 = 18 ÷ 2 = 9 (from eq.4a) F = (E + E²) ÷ (E – 1) = (3 + 3²) ÷ (3 – 1) = (3 + 9) ÷ 2 = 12 ÷ 2 = 6 (from eq.6b)
Solution
Substitute 3 for E, 6 for F, and 9 for D in eq.3: (C × 3) – 6 = 9 Add 6 to both sides of the above equation: (C × 3) – 6 + 6 = 9 + 6 which becomes 3×C = 15 Divide both sides of the equation above by 3: 3×C ÷ 3 = 15 ÷ 3 which makes C = 5 and makes ABCDEF = 475936