Puzzle for April 13, 2019 ( )
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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 B to both sides of eq.3: A - B + B = B + C + B which becomes eq.3a) A = 2×B + C
Hint #2
Add B to both sides of eq.4: B + C + E + B = A - B + F + B which becomes 2×B + C + E = A + F In the above equation, replace 2×B + C with A (from eq.3a): A + E = A + F Subtract A from both sides: A + E - A = A + F - A which makes E = F
Hint #3
In eq.6, replace F with E: D - B = E ÷ E which makes D - B = 1 (assumes E ≠ 0) Add B to each side of the equation above: D - B + B = 1 + B which means eq.6a) D = 1 + B
Hint #4
In eq.2, substitute 1 + B for D (from eq.6a): C = B + 1 + B which makes eq.2a) C = 2×B + 1
Hint #5
Substitute 2×B + 1 for C (from eq.2a) in eq.3a: A = 2×B + 2×B + 1 which makes eq.3b) A = 4×B + 1
Hint #6
Substitute (1 + B) for D (from eq.6a) in eq.5: E = (1 + B) × (1 + B) which becomes E = 1 + 2×B + B² and which means eq.5a) F = E = 1 + 2×B + B²
Hint #7
Substitute 4×B + 1 for A (from eq.3b), 2×B + 1 for C (from eq.2a), 1 + B for D (from eq.6a), and 1 + 2×B + B² for both E and F (from eq.5a): 4×B + 1 + B + 2×B + 1 + 1 + B + 1 + 2×B + B² + 1 + 2×B + B² = 37 which simplifies to 12×B + 5 + 2×B² = 37 Subtract 37 from each side: 12×B + 5 + 2×B² - 37 = 37 - 37 which becomes 12×B - 32 + 2×B² = 0 which may be written as 2×B² + 12×B - 32 = 0 Divide both sides by 2: (2×B² + 12×B - 32) ÷ 2 = 0 ÷ 2 which means eq.1a) B² + 6×B - 16 = 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)×6 ± sq.rt.[6² - (4 × 1 × (-16))]} ÷ (2 × 1) which becomes B = {-6 ± sq.rt.[36 - (-64)]} ÷ 2 which is equivalent to B = {-6 ± sq.rt.[36 + 64]} ÷ 2 which becomes B = {-6 ± sq.rt.[100]} ÷ 2 which becomes B = {-6 ± 10} ÷ 2 In the above equation, either: B = {-6 + 10} ÷ 2 = 4 ÷ 2 = 2 or: B = {-6 - 10} ÷ 2 = -16 ÷ 2 = -8 Since B must be a non-negative integer, then B ≠ -8 and therefore makes B = 2 making A = 4×B + 1 = 4×2 + 1 = 8 + 1 = 9 (from eq.3b) C = 2×B + 1 = 2×2 + 1 = 4 + 1 = 5 (from eq.2a) D = 1 + B = 1 + 2 = 3 (from eq.6a) E = F = 1 + 2×B + B² = 1 + 2×2 + 2² = 1 + 4 + 4 = 9 (from eq.5a) and ABCDEF = 925399