Puzzle for March 15, 2020 ( )
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 is a 2-digit number (not A×B).
Scratchpad
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Hint #1
In eq.3, replace D with B + F (from eq.2): E = B + F – F which makes E = B
Hint #2
eq.4 may be re-written as: C – A + D = B – C Subtract D from both sides of the above equation: C – A + D – D = B – C – D which becomes eq.4a) C – A = B – C – D
Hint #3
In eq.5, replace C – A with B – C – D (from eq.4a): D – C – E = B – C – D Add C, D, and E to both sides of the above equation: D – C – E + C + D + E = B – C – D + C + D + E which becomes eq.5a) 2×D = B + E
Hint #4
In eq.5a, substitute B for E: 2×D = B + B which makes 2×D = 2×B Divide both sides of the above equation by 2: 2×D ÷ 2 = 2×B ÷ 2 which makes D = B
Hint #5
Substitute B for D in eq.2: B + F = B Subtract B from both sides of the equation above: B + F – B = B – B which makes F = 0
Hint #6
Substitute B for both D and E in eq.6: A + (B × B) = 10×A + B Subtract A and B from each side of the equation above: A + (B × B) – A – B = 10×A + B – A – B which becomes B² – B = 9×A Divide both sides by 9: (B² – B) ÷ 9 = 9×A ÷ 9 which becomes eq.6a) (B² – B) ÷ 9 = A
Hint #7
Substitute B for D and E, and ((B² – B) ÷ 9) for A in eq.5: B – C – B = C – ((B² – B) ÷ 9) which becomes –C = C – ((B² – B) ÷ 9) Subtract C from both sides of the above equation: –C – C = C – ((B² – B) ÷ 9) – C which becomes –2×C = –((B² – B) ÷ 9) Divide both sides by (–2): –2×C ÷ (–2) = –((B² – B) ÷ 9) ÷ (–2) which becomes eq.5b) C = (B² – B) ÷ 18
Hint #8
Substitute (B² – B) ÷ 9 for A (from eq.6a), (B² – B) ÷ 18 for C (from eq.5b), B for D and E, and 0 for F in eq.1: (B² – B) ÷ 9 + B + (B² – B) ÷ 18 + B + B + 0 = 39 which becomes (B² – B) ÷ 9 + (B² – B) ÷ 18 + 3×B = 39 Multiply both sides of the equation above by 18: ((B² – B) ÷ 9 + (B² – B) ÷ 18 + 3×B) × 18 = 39 × 18 which becomes 2×B² – 2×B + B² – B + 54×B = 702 which becomes 3×B² + 51×B = 702 Divide both sides by 3: (3×B² + 51×B) ÷ 3 = 702 ÷ 3 which becomes B² + 17×B = 234 Subtract 234 from both sides: B² + 17×B – 234 = 234 – 234 which becomes eq.1a) B² + 17×B – 234 = 0
Solution
eq.1a is a quadratic equation in standard form. Using the quadratic equation solution formula to solve for B in eq.1a yields: B = {–17 ± sq.rt.(17² – (4 × 1 × (–234)))} ÷ (2 × 1) which becomes B = {–17 ± sq.rt.(289 + 936)} ÷ 2 which becomes B = {–17 ± sq.rt.(1225)} ÷ 2 B = (–17 ± 35) ÷ 2 In the above equation, either B = {–17 + 35} ÷ 2 = 18 ÷ 2 = 9 or B = {–17 – 35} ÷ 2 = –52 ÷ 2 = –26 Since B must be a non-negative integer, then B ≠ –26 and therefore makes B = 9 making A = (B² – B) ÷ 9 = (9² – 9) ÷ 9 = (81 – 9) ÷ 9 = 72 ÷ 9 = 8 (from eq.6a) C = (B² – B) ÷ 18 = (9² – 9) ÷ 18 = (81 – 9) ÷ 18 = 72 ÷ 18 = 4 (from eq.5b) D = E = B = 9 and ABCDEF = 894990