Puzzle for July 9, 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 non-negative integer.
* DE is a 2-digit number (not D×E).
Scratchpad
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Hint #1
Add C to both sides of eq.4: C - D + C = F - C + C which becomes eq.4a) 2×C - D = F
Hint #2
In eq.3, replace F with 2×C - D (from eq.4a): D + 2×C - D = A + C which becomes 2×C = A + C Subtract C from each side of the equation above: 2×C - C = A + C - C which makes A = C
Hint #3
In eq.2, replace F with 2×C - D (from eq.4a): 2×C - D = B + C + D Subtract C and D from both sides of the above equation: 2×C - D - C - D = B + C + D - C - D which becomes eq.2a) C - 2×D = B
Hint #4
In eq.5, substitute C for A, and C - 2×D for B (from eq.2a): D + E = C + C - 2×D which becomes D + E = 2×C - 2×D Subtract D from each side of the above equation: D + E - D = 2×C - 2×D - D which becomes eq.5a) E = 2×C - 3×D
Hint #5
eq.6 may be written as: eq.6a) A × C = B + 10×D + E
Hint #6
Substitute C for A, C - 2×D for B (from eq.2a), and 2×C - 3×D for E (from eq.5a) into eq.6a: C × C = C - 2×D + 10×D + 2×C - 3×D which becomes C² = 3×C + 5×D Subtract 3×C both sides of the equation above: C² - 3×C = 3×C + 5×D - 3×C which becomes C² - 3×C = 5×D Divide both sides by 5: (C² - 3×C) ÷ 5 = D which becomes eq.6b) ⅕×C² - ⅗×C = D
Hint #7
Substitute (⅕×C² - ⅗×C) for D (from eq.6b) in eq.2a: C - 2×(⅕×C² - ⅗×C) = B which becomes C - ⅖×C² + 1⅕×C = B which becomes eq.2b) 2⅕×C - ⅖×C² = B
Hint #8
Substitute (⅕×C² - ⅗×C) for D (from eq.6b) in eq.5a: E = 2×C - 3×(⅕×C² - ⅗×C) which becomes E = 2×C - ⅗×C² + 1⅘×C which becomes eq.5b) E = 3⅘×C - ⅗×C²
Hint #9
Substitute (⅕×C² - ⅗×C) for D (from eq.6b) in eq.4a: 2×C - (⅕×C² - ⅗×C) = F which becomes 2×C - ⅕×C² + ⅗×C = F which becomes eq.4b) 2⅗×C - ⅕×C² = F
Hint #10
Substitute C for A, 2⅕×C - ⅖×C² for B (from eq.2b), ⅕×C² - ⅗×C for D (from eq.6b), 3⅘×C - ⅗×C² for E (from eq.5b), and 2⅗×C - ⅕×C² for F (from eq.4b) in eq.1: C + 2⅕×C - ⅖×C² + C + ⅕×C² - ⅗×C + 3⅘×C - ⅗×C² + 2⅗×C - ⅕×C² = 25 which simplifies to 10×C - C² = 25 In the above equation, subtract 10×C from both sides, and add C² to both sides: 10×C - C² - 10×C + C² = 25 - 10×C + C² which becomes 0 = 25 - 10×C + C² which may be written as eq.1a) 0 = C² - 10×C + 25
Hint #11
eq.1a is a quadratic equation in standard form. The quadratic equation solution formula could be used to solve for C in eq.1a. However, eq.1a can be easily factored into the product of two expressions: 0 = (C - 5) × (C - 5) which makes C = 5
Solution
Since C = 5, then: A = C = 5 B = 2⅕×C - ⅖×C² = 2⅕×5 - ⅖×5² = 11 - ⅖×25 = 11 - 10 = 1 (from eq.2b) D = ⅕×C² - ⅗×C = ⅕×5² - ⅗×5 = ⅕×25 - 3 = 5 - 3 = 2 (from eq.6b) E = 3⅘×C - ⅗×C² = 3⅘×5 - ⅗×5² = 19 - ⅗×25 = 19 - 15 = 4 (from eq.5b) F = 2⅗×C - ⅕×C² = 2⅗×5 - ⅕×5² = 13 - ⅕×25 = 13 - 5 = 8 (from eq.4b) and ABCDEF = 515248