Puzzle for September 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.
* "D ^ F" means "D raised to the power of F".
Scratchpad
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Hint #1
Add F to both sides of eq.1: C - F + F = A + F + F which becomes eq.1a) C = A + 2×F
Hint #2
In eq.4, replace C with A + 2×F (from eq.1a): A + D + F = B + A + 2×F - D - F which becomes A + D + F = B + A + F - D In the above equation, subtract A and F from both sides, and add D to both sides: A + D + F - A - F + D = B + A + F - D - A - F + D which simplifies to 2×D = B
Hint #3
In eq.5, replace B with 2×D: 2×D = D × F Divide both sides of the above equation by D: 2×D ÷ D = (D × F) ÷ D which makes 2 = F
Hint #4
Subtract the left and right sides of eq.3 from the left and right sides from eq.2, respectively: B + D - (D + E) = A + E - (A + C) which becomes B + D - D - E = A + E - A - C which becomes B - E = E - C Add E and C to both sides of the above equation: B - E + E + C = E - C + E + C which becomes eq.2a) B + C = 2×E
Hint #5
In eq.6, substitute 2×E for B + C (from eq.2a), and 2 for F: 2×E = D ^ 2 which may be written as 2×E = D² Divide both sides of the above equation by 2: 2×E ÷ 2 = D² ÷ 2 which makes E = ½×D²
Hint #6
Substitute 2×D for B, and (½×D²) for E in eq.2a: 2×D + C = 2×(½×D²) which becomes 2×D + C = D² Subtract 2×D from each side of the equation above: 2×D + C - 2×D = D² - 2×D which becomes eq.2b) C = D² - 2×D
Hint #7
Substitute D² - 2×D for C (from eq.2b), and 2 for F in eq.1a: D² - 2×D = A + 2×2 which becomes D² - 2×D = A + 4 Subtract 4 from each side of the above equation: D² - 2×D - 4 = A + 4 - 4 which becomes eq.1b) D² - 2×D - 4 = A
Hint #8
Substitute ½×D² for E, D² - 2×D - 4 for A (from eq.1b), and D² - 2×D for C (from eq.2b) in eq.3: D + ½×D² = D² - 2×D - 4 + D² - 2×D which becomes D + ½×D² = 2×D² - 4×D - 4 Subtract D and ½×D² from both sides of the equation above: D + ½×D² - D - ½×D² = 2×D² - 4×D - 4 - D - ½×D² which becomes eq.3a) 0 = 1½×D² - 5×D - 4
Hint #9
eq.3a is a quadratic equation in standard form. Using the quadratic equation solution formula to solve for D in eq.3a yields: D = { (-1)×(-5) ± sq.rt.[(-5)² - (4 × 1½ × (-4))] } ÷ (2 × 1½) which becomes D = {5 ± sq.rt.(25 - (-24))} ÷ 3 which becomes D = {5 ± sq.rt.(49)} ÷ 3 which becomes eq.3b) D = (5 ± 7) ÷ 3
Hint #10
In eq.3b, either: D = (5 + 7) ÷ 3 = 12 ÷ 3 = 4 or: D = (5 - 7) ÷ 3 = -2 ÷ 3 = -⅔ Since D must be a positive integer, then D ≠ -⅔ and therefore makes D = 4
Solution
Since D = 4, then: A = D² - 2×D - 4 = 4² - 2×4 - 4 = 16 - 8 - 4 = 4 (from eq.1b) B = 2×D = 2 × 4 = 8 C = D² - 2×D = 4² - 2×4 = 16 - 8 = 8 (from eq.2b) E = ½×D² = ½×(4²) = ½×16 = 8 and ABCDEF = 488482