Puzzle for July 10, 2022 ( )
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.
* AB and BC are 2-digit numbers (not A×B or B×C).
** "F^D" means "F raised to the power of D".
Scratchpad
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Hint #1
Add D and F to both sides of eq.3: B – D + E + D + F = A + C – F + D + F which becomes eq.3a) B + E + F = A + C + D In eq.3a, replace E + F with C + D (from eq.2): A + C + D = B + C + D Subtract C and D from both sides: A + C + D – C – D = B + C + D – C – D which makes A = B
Hint #2
eq.4 may be written as: 10×A + B + C + D = A + 10×B + C – D In the above equation, subtract A and B and C from both sides, and add D to both sides: 10×A + B + C + D – A – B – C + D = A + 10×B + C – D – A – B – C + D which simplifies to eq.4a) 9×A + 2×D = 9×B
Hint #3
In eq.4a, replace B with A: 9×A + 2×D = 9×A Subtract 9×A from each side of the equation above: 9×A + 2×D – 9×A = 9×A – 9×A which makes 2×D = 0 which means D = 0
Hint #4
eq.5 may be written as: E = (A + B + C + F) ÷ 4 Multiply both sides of the above equation by 4: 4 × E = 4 × (A + B + C + F) ÷ 4 which becomes eq.5a) 4×E = A + B + C + F
Hint #5
eq.1 may be written as: A + B + C + F + D + E = 30 In the equation above, substitute 4×E for A + B + C + F (from eq.5a), and 0 for D: 4×E + 0 + E = 30 which becomes 5×E = 30 Divide both sides by 5: 5×E ÷ 5 = 30 ÷ 5 which makes E = 6
Hint #6
Substitute 0 for D, and 6 for E in eq.2: C + 0 = 6 + F which makes eq.2a) C = 6 + F
Hint #7
Substitute 0 for D, A for B, and 6 for E in eq.6: F^0 = A×A – C×6 which becomes eq.6a) 1 = A² – 6×C (implies F ≠ 0)
Hint #8
In eq.3a, substitute 0 for D: B + E + F = A + C + 0 which becomes eq.3b) B + E + F = A + C
Hint #9
eq.1 may be written as: A + C + D + B + E + F = 30 In the above equation, substitute 0 for D, and A + C for B + E + F (from eq.3b): A + C + 0 + A + C = 30 which becomes 2×A + 2×C = 30 Divide both sides by 2: (2×A + 2×C) ÷ 2 = 30 ÷ 2 which becomes A + C = 15 Subtract A from both sides: A + C – A = 15 – A which makes eq.1a) C = 15 – A
Hint #10
Substitute (15 – A) for C (from eq.1a) into eq.6a: 1 = A² – 6×(15 – A) which becomes 1 = A² – 90 + 6×A Subtract 1 from each side of the equation above: 1 – 1 = A² – 90 + 6×A – 1 which becomes 0 = A² – 91 + 6×A which may be written as eq.6b) 0 = A² + 6×A – 91
Hint #11
eq.6b is a quadratic equation in standard form. Using the quadratic equation solution formula to solve for A in eq.6b yields: A = {(–1)×(6) ± sq.rt.[(6)² – (4 × (1) × (–91))]} ÷ (2 × (1)) which becomes A = {–6 ± sq.rt.(36 – (–364)} ÷ 2 which becomes A = (–6 ± sq.rt.(400)) ÷ 2 which becomes A = (–6 ± 20) ÷ 2 In the above equation, either A = (–6 + 20) ÷ 2 = 14 ÷ 2 = 7 or A = (–6 – 20) ÷ 2 = –26 ÷ 2 = –13 Since A must be a non-negative one-digit integer, then A ≠ –13 and therefore makes A = 7 and also makes B = A = 7
Hint #12
Substitute 7 for A in eq.1a: C = 15 – 7 which makes C = 8
Solution
Substitute 8 for C in eq.2a: 8 = 6 + F Subtract 6 from each side of the above equation: 8 – 6 = 6 + F – 6 which makes 2 = F and makes ABCDEF = 778062