Puzzle for May 16, 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.
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Hint #1
eq.7 may be written as an exponential equation: A ^ E = A × B ("A ^ E" means "A raised to the power of E") Since eq.5 and eq.6 and eq.7 imply that A ≠ 0, divide both sides of the above equation by A: (A ^ E) ÷ A = A × B ÷ A which becomes eq.7a) A ^ (E - 1) = B
Hint #2
eq.7a may be re-written as a logarithmic equation: log [base A] (B) = E - 1 In the above equation, replace log [base A] (B) with D (from eq.6): eq.7b) D = E - 1
Hint #3
In eq.4, replace D with E - 1 (from eq.7b): A + F = E - 1 + E which becomes A + F = 2×E - 1 Subtract A from each side of the equation above: A + F - A = 2×E - 1 - A which becomes eq.4a) F = 2×E - 1 - A
Hint #4
In eq.3, substitute E - 1 for D (from eq.7b), and 2×E - 1 - A for F (from eq.4a): E - 1 + E + 2×E - 1 - A - A = A + B which becomes 4×E - 2 - 2×A = A + B Subtract A from both sides of the equation above: 4×E - 2 - 2×A - A = A + B - A which becomes eq.3a) 4×E - 2 - 3×A = B
Hint #5
Substitute 2×E - 1 - A for F (from eq.4a), and E - 1 for D (from eq.7b) in eq.2: C + 2×E - 1 - A = A - C + E - 1 Add 1, A, and C to both sides of the above equation: C + 2×E - 1 - A + 1 + A + C = A - C + E - 1 + 1 + A + C which simplifies to 2×C + 2×E = 2×A + E Subtract 2×E from both sides: 2×C + 2×E - 2×E = 2×A + E - 2×E which becomes 2×C = 2×A - E Divide both sides by 2: 2×C ÷ 2 = (2×A - E) ÷ 2 which becomes eq.2a) C = A - ½×E
Hint #6
In eq.1, substitute 4×E - 2 - 3×A for B (from eq.3a), A - ½×E for C (from eq.2a), E - 1 for D (from eq.7b), and 2×E - 1 - A for F (from eq.4a): A + 4×E - 2 - 3×A + A - ½×E + E - 1 + E + 2×E - 1 - A = 22 which simplifies to 7½×E - 4 - 2×A = 22 Add 4 and 2×A to both sides of the above equation: 7½×E - 4 - 2×A + 4 + 2×A = 22 + 4 + 2×A which becomes eq.1a) 7½×E = 26 + 2×A
Hint #7
Subtract A and D from both sides of eq.4: A + F - A - D = D + E - A - D which becomes F - D = E - A Substitute E ÷ A for F - D (from eq.5) into the equation above: eq.4b) E ÷ A = E - A
Hint #8
In eq.4b, add A to both sides, and subtract (E ÷ A) from both sides: (E ÷ A) + A - (E ÷ A) = E - A + A - (E ÷ A) which becomes A = E - (E ÷ A) which may be written as A = E × (1 - (1 ÷ A)) Divide both sides of the above equation by (1 - (1 ÷ A)): A ÷ (1 - (1 ÷ A)) = E × (1 - (1 ÷ A)) ÷ (1 - (1 ÷ A)) which becomes eq.4c) A ÷ (1 - (1 ÷ A)) = E
Hint #9
Substitute (A ÷ A) for the first 1 in eq.4c: A ÷ ((A ÷ A) - (1 ÷ A)) = E which is the same as A ÷ ((A - 1) ÷ A)) = E which is equivalent to eq.4d) A² ÷ (A - 1) = E
Hint #10
Substitute (A² ÷ (A - 1)) for E (from eq.4d) in eq.1a: 7½×(A² ÷ (A - 1)) = 26 + 2×A Multiply both sides of the equation above by (A - 1): (7½×A² ÷ (A - 1)) × (A - 1) = (26 + 2×A) × (A - 1) which becomes 7½×A² = 26×A + 2×A² - 26 - 2×A which becomes 7½×A² = 24×A + 2×A² - 26 Subtract 7½×A² from each side of the equation above: 7½×A² - 7½×A² = 24×A + 2×A² - 26 - 7½×A² which becomes 0 = 24×A - 5½×A² - 26 which may be written as eq.1b) 0 = -5½×A² + 24×A - 26
Hint #11
eq.1b is a quadratic equation in standard form. Using the quadratic equation solution formula to solve for A in eq.1b yields: A = { (-1)×(24) ± sq.rt.[(24)² - (4 × (-5½) × (-26))] } ÷ (2 × (-5½)) which becomes A = {-24 ± sq.rt.(576 - 572)} ÷ (-11) which becomes A = {-24 ± sq.rt.(4)} ÷ (-11) which becomes A = (-24 ± 2) ÷ (-11) In the above equation, either A = (-24 + 2) ÷ (-11) = -22 ÷ (-11) = 2 or A = (-24 - 2) ÷ (-11) = -26 ÷ (-11) = 2.363636363636 Since A must be an integer, then A ≠ 2.363636363636 and therefore makes A = 2
Solution
Substitute 2 for A in eq.4d: 2² ÷ (2 - 1) = E which becomes 4 ÷ (1) = E which means 4 = E making B = A ^ (E - 1) = 2 ^ (4 - 1) = 2 ^ (3) = 8 (from eq.7a) C = A - ½×E = 2 - ½×4 = 2 - 2 = 0 (from eq.2a) D = E - 1 = 4 - 1 = 3 (from eq.7b) F = 2×E - 1 - A = 2×4 - 1 - 2 = 8 - 3 = 5 (from eq.4a) and ABCDEF = 280345