Puzzle for September 12, 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
Add F to both sides of eq.3: E – F + F = C + F + F which becomes E = C + 2×F In eq.5, replace E with C + 2×F: A + C + 2×F = C + D + F Subtract C and F from each side of the above equation: A + C + 2×F – C – F = C + D + F – C – F which simplifies to eq.5a) A + F = D
Hint #2
Add F to both sides of eq.2: C – F + F = A + F which becomes eq.2a) C = A + F In eq.2a, replace A + F with D (from eq.5a): C = D
Hint #3
In eq.4, substitute A + F for C (from eq.2a): A + F – A = A – B which becomes F = A – B Add B to both sides of the above equation: F + B = A – B + B which becomes F + B = A which is the same as eq.4a) B + F = A
Hint #4
Substitute A for B + F (from eq.4a) in eq.6: A = E ÷ A Multiply both sides of the equation above by A: A × A = (E ÷ A) × A which makes A² = E
Hint #5
In eq.3, replace E with A², and C with A + F (from eq.2a): A² – F = A + F + F which becomes A² – F = A + 2×F In the above equation, add F to both sides, and subtract A from both sides: A² – F + F – A = A + 2×F + F – A which becomes A² – A = 3×F Divide both sides by 3: (A² – A) ÷ 3 = 3×F ÷ 3 which becomes eq.3a) (A² – A) ÷ 3 = F
Hint #6
Subtract F from both sides of eq.4a: B + F – F = A – F which becomes eq.4b) B = A – F In eq.4b, substitute ((A² – A) ÷ 3) for F (from eq.3a): eq.4c) B = A – ((A² – A) ÷ 3)
Hint #7
In eq.5a, substitute ((A² – A) ÷ 3) for F (from eq.3a): A + ((A² – A) ÷ 3) = D which also makes eq.5b) C = D = A + ((A² – A) ÷ 3)
Hint #8
In eq.1, substitute –– A – ((A² – A) ÷ 3) for B (from eq.4c), A + ((A² – A) ÷ 3) for C and D (from eq.5b), A² for E, and ((A² – A) ÷ 3) for F (from eq.3a): A + A – ((A² – A) ÷ 3) + A + ((A² – A) ÷ 3) + A + ((A² – A) ÷ 3) + A² + ((A² – A) ÷ 3) = 25 which becomes 4×A + A² + 2×((A² – A) ÷ 3) = 25 which becomes 4×A + A² + ⅔×A² – ⅔×A = 25 which becomes 3⅓×A + 1⅔×A² = 25 Subtract 25 from each side of the above equation: 3⅓×A + 1⅔×A² – 25 = 25 – 25 which is equivalent to eq.1a) 1⅔×A² + 3⅓×A – 25 = 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)×(3⅓) ± sq.rt.[(3⅓)² – (4 × (1⅔) × (–25))] } ÷ (2 × (1⅔)) which becomes A = {–3⅓ ± sq.rt.(100/9 – (–500/3))} ÷ 3⅓ which becomes A = (–3⅓ ± sq.rt.(100/9 + 1500/9)) ÷ 3⅓ which becomes A = (–3⅓ ± sq.rt.(1600/9)) ÷ 3⅓ which becomes A = (–3⅓ ± 400/3) ÷ 3⅓ which becomes A = (–3⅓ ± 13⅓) ÷ 3⅓ In the above equation, either A = (–3⅓ + 13⅓) ÷ 3⅓ = 10 ÷ 3⅓ = 3 or A = (–3⅓ – 13⅓) ÷ 3⅓ = –16⅔ ÷ 3⅓ = –5 Since A must be non-negative, then A ≠ –5 and therefore makes A = 3
Solution
Substitute 3 for A in eq.3a: (3² – 3) ÷ 3 = F which becomes (9 – 3) ÷ 3 = F which becomes 6 ÷ 3 = F which means 2 = F making B = A – F = 3 – 2 = 1 (from eq.4b) C = A + F = 3 + 2 = 5 (from eq.2a) D = C = 5 E = A² = 3² = 9 and ABCDEF = 315592