Puzzle for May 14, 2023 ( )
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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 non-negative integer.
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Hint #1
In eq.4, replace D with A + F (from eq.5): B = A + F + F which becomes eq.4a) B = A + 2×F
Hint #2
In eq.3, replace B with A + 2×F (from eq.4a): C = A + 2×F + F which becomes eq.3a) C = A + 3×F
Hint #3
In eq.2, replace C with A + 3×F (from eq.3a): E = A + 3×F + F which becomes eq.2a) E = A + 4×F
Hint #4
In eq.1, substitute A + 2×F for B (from eq.4a), A + 3×F for C (from eq.3a), A + F for D (from eq.5), and A + 4×F for E (from eq.2a): A + A + 2×F + A + 3×F + A + F + A + 4×F + F = 36 which becomes eq.1a) 5×A + 11×F = 36
Hint #5
In eq.6, substitute A + 4×F for E (from eq.2a), (A + F) for D (from eq.5), A + 2×F for B (from eq.4a), and A + 3×F for C (from eq.3a): A + 4×F - (A + F) = (A + 2×F + A + 3×F) ÷ A which becomes A + 4×F - A - F = (2×A + 5×F) ÷ A which becomes eq.6a) 3×F = (2×A + 5×F) ÷ A
Hint #6
Multiply both sides of eq.6a by A: A × 3×F = A × (2×A + 5×F) ÷ A which becomes A × 3×F = 2×A + 5×F Subtract 2×A from each side of the equation above: A × 3×F - 2×A = 2×A + 5×F - 2×A which becomes A × 3×F - 2×A = 5×F which may be written as A × (3×F - 2) = 5×F Divide both sides by (3×F - 2): (A × (3×F - 2)) ÷ (3×F - 2) = 5×F ÷ (3×F - 2) which becomes eq.6b) A = 5×F ÷ (3×F - 2)
Hint #7
Substitute (5×F ÷ (3×F - 2)) for A (from eq.6b) in eq.1a: 5×(5×F ÷ (3×F - 2)) + 11×F = 36 which becomes 25×F ÷ (3×F - 2) + 11×F = 36 Subtract 11×F from both sides of the equation above: 25×F ÷ (3×F - 2) + 11×F - 11×F = 36 - 11×F which becomes eq.1b) 25×F ÷ (3×F - 2) = 36 - 11×F
Hint #8
Multiply both sides of eq.1b by (3×F - 2): 25×F ÷ (3×F - 2) × (3×F - 2) = (36 - 11×F) × (3×F - 2) which becomes 25×F = 108×F - 33×F² - 72 + 22×F which becomes 25×F = 130×F - 33×F² - 72 Subtract 25×F from each side of the equation above: 25×F - 25×F = 130×F - 33×F² - 72 - 25×F which becomes 0 = 105×F - 33×F² - 72 which may be written as 0 = -33×F² + 105×F - 72 Divide both sides by 3: 0 ÷ 3 = (-33×F² + 105×F - 72) ÷ 3 which becomes eq.1c) 0 = -11×F² + 35×F - 24
Hint #9
eq.1c is a quadratic equation in standard form. Using the quadratic equation solution formula to solve for F in eq.1c yields: F = { (-1)×(35) ± sq.rt.[(35)² - (4 × (-11) × (-24))] } ÷ (2 × (-11)) which becomes F = {-35 ± sq.rt.(1225 - 1056)} ÷ (-22) which becomes F = {-35 ± sq.rt.(169)} ÷ (-22) which becomes F = (-35 ± 13) ÷ (-22) In the above equation, either: F = (-35 + 13) ÷ (-22) = -22 ÷ (-22) = 1 or: F = (-35 - 13) ÷ (-22) = -48 ÷ (-22) = 2.1818181818 Since F must be an integer, then F ≠ 2.1818181818 and therefore makes F = 1
Hint #10
Substitute 1 for F in eq.6b: A = 5×1 ÷ (3×1 - 2) which becomes A = 5 ÷ (3 - 2) which becomes A = 5 ÷ 1 which makes A = 5
Solution
Since F = 1 and A = 5, then: B = A + 2×F = 5 + 2×1 = 5 + 2 = 7 (from eq.4a) C = A + 3×F = 5 + 3×1 = 5 + 3 = 8 (from eq.3a) D = A + F = 5 + 1 = 6 (from eq.5) E = A + 4×F = 5 + 4×1 = 5 + 4 = 9 (from eq.2a) and ABCDEF = 578691