Puzzle for August 20, 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.
Scratchpad
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Hint #1
eq.5 may be written as: D = (B + C + E) ÷ 3 Multiply both sides of the above equation by 3: 3 × D = 3 × (B + C + E) ÷ 3 which becomes eq.5a) 3×D = B + C + E
Hint #2
Add A and D to both sides of eq.2: D - A + A + D = E - D + A + D which becomes eq.2a) 2×D = E + A eq.5a may be written as: eq.5b) 2×D + D = B + C + E
Hint #3
In eq.5b, replace 2×D with E + A (from eq.2a): E + A + D = B + C + E Subtract E from each side of the equation above: E + A + D - E = B + C + E - E which becomes eq.5c) A + D = B + C
Hint #4
In eq.3, replace A + D with B + C (from eq.5c): B + F = B + C Subtract B from each side of the equation above: B + F - B = B + C - B which makes F = C
Hint #5
eq.4 may be written as: C + D + F = E + A + B In the equation above, substitute C for F, and 2×D for E + A (from eq.2a): C + D + C = 2×D + B which becomes 2×C + D = 2×D + B Subtract D and B from both sides: 2×C + D - D - B = 2×D + B - D - B which becomes eq.4a) 2×C - B = D
Hint #6
Substitute (2×C - B) for D (from eq.4a) in eq.5a: 3×(2×C - B) = B + C + E which becomes 6×C - 3×B = B + C + E Subtract B and C from each side of the equation above: 6×C - 3×B - B - C = B + C + E - B - C which becomes eq.5d) 5×C - 4×B = E
Hint #7
Substitute 2×C - B for D (from eq.4a) in eq.5c: A + 2×C - B = B + C In the above equation, subtract 2×C from both sides, and add B to both sides: A + 2×C - B - 2×C + B = B + C - 2×C + B which becomes eq.5e) A = 2×B - C
Hint #8
Substitute 2×C - B for D (from eq.4a), 5×C - 4×B for E (from eq.5d), and C for F in eq.6: B × C = 2×C - B + 5×C - 4×B + C which becomes B × C = 8×C - 5×B Add 5×B to both sides of the above equation: B × C + 5×B = 8×C - 5×B + 5×B which becomes B × (C + 5) = 8×C Divide both sides by (C + 5): B × (C + 5) ÷ (C + 5) = 8×C ÷ (C + 5) which becomes eq.6a) B = 8×C ÷ (C + 5)
Hint #9
Substitute (8×C ÷ (C + 5)) for B (from eq.6a) in eq.5e: A = 2×(8×C ÷ (C + 5)) - C which becomes eq.5f) A = (16×C ÷ (C + 5)) - C
Hint #10
Substitute (8×C ÷ (C + 5)) for B (from eq.6a) in eq.4a: eq.4b) 2×C - (8×C ÷ (C + 5)) = D
Hint #11
Substitute (8×C ÷ (C + 5)) for B (from eq.6a) in eq.5d: 5×C - 4×(8×C ÷ (C + 5)) = E which becomes eq.5g) 5×C - (32×C ÷ (C + 5)) = E
Hint #12
In eq.1, substitute: (16×C ÷ (C + 5)) - C for A (from eq.5f), 8×C ÷ (C + 5) for B (from eq.6a), 2×C - (8×C ÷ (C + 5)) for D (from eq.4b), 5×C - (32×C ÷ (C + 5)) for E (from eq.5g), and C for F: (16×C ÷ (C + 5)) - C + 8×C ÷ (C + 5) + C + 2×C - (8×C ÷ (C + 5)) + 5×C - (32×C ÷ (C + 5)) + C = 32 which simplifies to 8×C - (16×C ÷ (C + 5)) = 32 Divide both sides of the above equation by 8: (8×C - (16×C ÷ (C + 5))) ÷ 8 = 32 ÷ 8 which becomes eq.1a) C - (2×C ÷ (C + 5)) = 4
Hint #13
In eq.1a, add (2×C ÷ (C + 5)) to both sides, and subtract 4 from both sides: C - (2×C ÷ (C + 5)) + (2×C ÷ (C + 5)) - 4 = 4 + (2×C ÷ (C + 5)) - 4 which becomes C - 4 = (2×C ÷ (C + 5)) Multiply both sides of the above equation by (C + 5): (C - 4) × (C + 5) = (2×C ÷ (C + 5)) × (C + 5) which becomes C² - 4×C + 5×C - 20 = 2×C which becomes C² + C - 20 = 2×C Subtract 2×C from each side: C² + C - 20 - 2×C = 2×C - 2×C which becomes eq.1b) C² - C - 20 = 0
Hint #14
eq.1b is a quadratic equation in standard form. The quadratic equation solution formula could be used to solve for C in eq.1b. However, eq.1b can easily be factored into the product of two expressions: C² - C - 20 = 0 may be written as (C + 4) × (C - 5) = 0 The above equation makes either: (C + 4) = 0 which would make C = -4 or: (C - 5) = 0 which would make C = 5 Since C is non-negative, then: C ≠ -4 and therefore makes C = 5
Solution
Since C = 5, then: A = (16×5 ÷ (5 + 5)) - 5 = (80 ÷ (10)) - 5 = 8 - 5 = 3 (from eq.5f) B = 8×5 ÷ (5 + 5) = 40 ÷ (10) = 4 (from eq.6a) D = 2×5 - (8×5 ÷ (5 + 5)) = 10 - (40 ÷ (10)) = 10 - (4) = 6 (from eq.4b) E = 5×5 - (32×5 ÷ (5 + 5)) = 25 - (160 ÷ (10)) = 25 - (16) = 9 (from eq.5g) F = 5 and ABCDEF = 345695