Puzzle for April 19, 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.
Scratchpad
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Hint #1
Add C and D to both sides of eq.3: A – C + C + D = B – D + C + D which becomes A + D = B + C In eq.2, replace B + C with A + D: D + E = A + D Subtract D from each side of the above equation: D + E – D = A + D – D which makes E = A
Hint #2
Add F and D to both sides of eq.4: E – F + F + D = C – D + F + D which becomes E + D = C + F which may be written as eq.4a) D + E = C + F In eq.4a above, replace D + E with B + C (from eq.2): B + C = C + F Subtract C from each side of the equation above: B + C – C = C + F – C which makes B = F
Hint #3
In eq.5, substitute C + F for D + E (from eq.4a): C + C + F = A + B + F which becomes 2×C + F = A + B + F Subtract F from both sides of the above equation: 2×C + F – F = A + B + F – F which becomes 2×C = A + B Divide both sides by 2: 2×C ÷ 2 = (A + B) ÷ 2 which becomes eq.5a) C = ½×A + ½×B
Hint #4
In eq.6, substitute B for F, ½×A + ½×B for C (from eq.5a), and A for E: B × B = A + ½×A + ½×B + A which becomes B² = 2½×A + ½×B Subtract ½×B from each side of the above equation: B² – ½×B = 2½×A + ½×B – ½×B which becomes eq.6a) B² – ½×B = 2½×A
Hint #5
Multiply both sides of eq.6a by 2 (to eliminate fractions): 2×(B² – ½×B) = 2×(2½×A) which makes 2×B² – B = 5×A Substitute E for A in the above equation: 2×B² – B = 5×E which means eq.6b) 2×B² – B = 5×E = 5×A
Hint #6
Multiply both sides of eq.5a by 10: 10×C = 10×(½×A + ½×B) which makes 10×C = 5×A + 5×B In the above equation, substitute 2×B² – B for 5×A (from eq.6b): 10×C = 2×B² – B + 5×B which makes eq.5b) 10×C = 2×B² + 4×B
Hint #7
Multiply both sides of eq.2 by 10: 10×(D + E) = 10×(B + C) which becomes 10×D + 10×E = 10×B + 10×C which may be written as 10×D + 2×(5×E) = 10×B + 10×C Substitute 2×B² – B for 5×E (from eq.6b), and 2×B² + 4×B for 10×C (from eq.5b) in the above equation: 10×D + 2×(2×B² – B) = 10×B + 2×B² + 4×B which becomes 10×D + 4×B² – 2×B = 2×B² + 14×B Add 2×B to both sides, and subtract 4×B² from both sides: 10×D + 4×B² – 2×B + 2×B – 4×B² = 2×B² + 14×B + 2×B – 4×B² which becomes eq.2a) 10×D = 16×B – 2×B²
Hint #8
Multiply both sides of eq.1 by 10: 10×(A + B + C + D + E + F) = 10×38 which becomes 10×A + 10×B + 10×C + 10×D + 10×E + 10×F = 380 which is equivalent to 2×(5×A) + 10×B + 10×C + 10×D + 2×(5×E) + 10×F = 380 Substitute 2×B² – B for 5×A and 5×E (from eq.6b), 2×B² + 4×B for 10×C (from eq.5b), 16×B – 2×B² for 10×D (from eq.2a), and B for F in the above equation: 2×(2×B² – B) + 10×B + 2×B² + 4×B + 16×B – 2×B² + 2×(2×B² – B) + 10×B = 380 which becomes 4×B² – 2×B + 40×B + 4×B² – 2×B = 380 which becomes 8×B² + 36×B = 380 Divide both sides by 4: (8×B² + 36×B) ÷ 4 = 380 ÷ 4 which becomes 2×B² + 9×B = 95 Subtract 95 from both sides: 2×B² + 9×B – 95 = 95 – 95 which means eq.1a) 2×B² + 9×B – 95 = 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 = { (–1)×(9) ± sq.rt.[(9)² – (4 × (2) × (–95))] } ÷ (2 × 2) which becomes B = {–9 ± sq.rt.(81 – (–760))} ÷ 4 which becomes B = {–9 ± sq.rt.(81 + 760)} ÷ 4 which becomes B = {–9 ± sq.rt.(841)} ÷ 4 which becomes B = (–9 ± 29) ÷ 4 In the equation above, either: B = (–9 + 29) ÷ 4 = 20 ÷ 4 = 5 or B = (–9 – 29) ÷ 4 = –38 ÷ 4 = –9½ Since B must be a non-negative integer, then B ≠ –9½ and therefore makes B = 5 making 5×A = 5×E = 2×B² – B = 2×5² – 5 = 2×25 – 5 = 50 – 5 = 45 which makes A = E = 45 ÷ 5 = 9 (from eq.6b) 10×C = 2×B² + 4×B = 2×5² + 4×5 = 2×25 + 20 = 50 + 20 = 70 which makes C = 70 ÷ 10 = 7 (from eq.5b) 10×D = –2×B² + 16×B = –2×5² + 16×5 = –2×25 + 80 = –50 + 80 = 30 which makes D = 30 ÷ 10 = 3 (from eq.2a) F = B = 5 and ABCDEF = 957395