Puzzle for December 17, 2023 ( )
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
In eq.5, replace B with C + F (from eq.2): D + F = A + C + F + C which becomes D + F = A + 2×C + F Subtract F from each side of the equation above: D + F - F = A + 2×C + F - F which becomes eq.5a) D = A + 2×C
Hint #2
In eq.4, replace D with A + 2×C (from eq.5a): A + A + 2×C = E - C which becomes 2×A + 2×C = E - C Add C to both sides of the above equation: 2×A + 2×C + C = E - C + C which becomes eq.4a) 2×A + 3×C = E
Hint #3
In eq.3, substitute 2×A + 3×C for E (from eq.4a), and (A + 2×C) for D (from eq.5a): 2×A + 3×C - (A + 2×C) = (A + 2×C) - A which becomes 2×A + 3×C - A - 2×C = 2×C which becomes A + C = 2×C Subtract C from both sides of the equation above: A + C - C = 2×C - C which becomes A = C
Hint #4
Substitute A for C in eq.4a: 2×A + 3×A = E which makes 5×A = E
Hint #5
Substitute A for C in eq.5a: D = A + 2×A which makes D = 3×A
Hint #6
Substitute 3×A for D, 5×A for E, and B for C + F (from eq.2) in eq.6: 3×A × 5×A = B + B + 3×A which becomes 15×A² = 2×B + 3×A Subtract 3×A from each sides of the above equation: 15×A² - 3×A = 2×B + 3×A - 3×A which becomes 15×A² - 3×A = 2×B Divide both sides by 2: (15×A² - 3×A) ÷ 2 = 2×B ÷ 2 which becomes eq.6a) 7½×A² - 1½×A = B
Hint #7
Substitute 7½×A² - 1½×A for B (from eq.6a), and A for C in eq.2: 7½×A² - 1½×A = A + F Subtract A from each side of the above equation: 7½×A² - 1½×A - A = A + F - A which becomes eq.2a) 7½×A² - 2½×A = F
Hint #8
Substitute 7½×A² - 1½×A for B (from eq.6a), A for C, 3×A for D, 5×A for E, and 7½×A² - 2½×A for F (from eq.2a) in eq.1: A + 7½×A² - 1½×A + A + 3×A + 5×A + 7½×A² - 2½×A = 21 which simplifies to 6×A + 15×A² = 21 Subtract 21 from both sides of the equation above: 6×A + 15×A² - 21 = 21 - 21 which becomes 6×A + 15×A² - 21 = 0 which may be written as 15×A² + 6×A - 21 = 0 Divide both sides by 3: (15×A² + 6×A - 21) ÷ 3 = 0 ÷ 3 which becomes eq.1a) 5×A² + 2×A - 7 = 0
Hint #9
eq.1a is a quadratic equation in standard form. Using the quadratic equation solution formula to solve for A in eq.1a yields: A = { (-1)×2 ± sq.rt.[2² - (4 × 5 × (-7))] } ÷ (2 × 5) which becomes A = {-2 ± sq.rt.(4 - (-140))} ÷ 10 which becomes A = {-2 ± sq.rt.(144)} ÷ 10 which becomes eq.1b) A = (-2 ± 12) ÷ 10
Hint #10
In eq.1b, either: A = (-2 + 12) ÷ 10 = 10 ÷ 10 = 1 or: A = (-2 - 12) ÷ 10 = -14 ÷ 10 = -1.4 Since A must be a non-negative integer, then A ≠ -1.4 and therefore makes A = 1
Solution
Since A = 1, then: B = 7½×A² - 1½×A = 7½×1² - 1½×1 = 7½×1 - 1½ = 7½ - 1½ = 6 (from eq.6a) C = A = 1 D = 3×A = 3×1 = 3 E = 5×A = 5×1 = 5 F = 7½×A² - 2½×A = 7½×1² - 2½×1 = 7½×1 - 2½ = 7½ - 2½ = 5 (from eq.2a) and ABCDEF = 161355