Puzzle for October 6, 2019 ( )
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
eq.1 may be re-written as: A + C + B + D + E + F = 29 In the above equation, replace A + C with B (from eq.3), and E + F with B (from eq.2): B + B + D + B = 29 which becomes 3×B + D = 29 Subtract 3×B from each side: 3×B + D – 3×B = 29 – 3×B which becomes eq.1a) D = 29 – 3×B
Hint #2
Add D to both sides of eq.5: B + C – D + D = D + D which becomes B + C = 2×D In the equation above, substitute (29 – 3×B) for D (from eq.1a): B + C = 2×(29 – 3×B) which becomes B + C = 58 – 6×B Subtract B from both sides: B + C – B = 58 – 6×B – B which becomes eq.5a) C = 58 – 7×B
Hint #3
In eq.3, substitute 58 – 7×B for C (from eq.5a): B = A + 58 – 7×B In the above equation, add 7×B to both sides, and subtract 58 from each side: B + 7×B – 58 = A + 58 – 7×B + 7×B – 58 which simplifies to eq.3a) 8×B – 58 = A
Hint #4
Substitute 8×B – 58 for A (from eq.3a), and 58 – 7×B for C (from eq.5a) in eq.4: 8×B – 58 + E = B + 58 – 7×B In the above equation, add 58 to both sides, and subtract 8×B from each side: 8×B – 58 + E + 58 – 8×B = B + 58 – 7×B + 58 – 8×B which simplifies to eq.4a) E = 116 – 14×B
Hint #5
Substitute 116 – 14×B for E (from eq.4a) in eq.2: 116 – 14×B + F = B Add 14×B to both sides, and subtract 116 from each side: 116 – 14×B + F + 14×B – 116 = B + 14×B – 116 which simplifies to eq.2a) F = 15×B – 116
Hint #6
Substitute (58 – 7×B) for C (from eq.5a), and (15×B – 116) for F (from eq.2a) in eq.6: (58 – 7×B) × (15×B – 116) = B which becomes 870×B – 6728 – 105×B² + 812×B = B Subtract B from each side of the equation above: 870×B – 6728 – 105×B² + 812×B – B = B – B which becomes –6728 – 105×B² + 1681×B = 0 which may be written as eq.6a) –105×B² + 1681×B – 6728 = 0
Solution
eq.6a is a quadratic equation in standard form. Using the quadratic equation solution formula to solve for B in eq.6a yields: B = { (–1)×(1681) ± sq.rt.[(1681)² – (4 × (–105) × (–6728))] } ÷ (2 × (–105)) which becomes B = {–1681 ± sq.rt.[2825761 – 2825760]} ÷ (–210) which becomes B = {–1681 ± sq.rt.[1]} ÷ (–210) which becomes B = {–1681 ± 1} ÷ (–210) In the above equation, either B = {–1681 + 1} ÷ (–210) = –1680 ÷ (–210) = 8 or B = {–1681 – 1} ÷ (–210) = –1682 ÷ (–210) = 8.0095238095238 Since B must be an integer, then B ≠ 8.0095238095238 Therefore, B = 8 making A = 8×B – 58 = 8×8 – 58 = 64 – 58 = 6 (from eq.3a) C = 58 – 7×B = 58 – 7×8 = 58 – 56 = 2 (from eq.5a) D = 29 – 3×B = 29 – 3×8 = 29 – 24 = 5 (from eq.1a) E = 116 – 14×B = 116 – 14×8 = 116 – 112 = 4 (from eq.4a) F = 15×B – 116 = 15×8 – 116 = 120 – 116 = 4 (from eq.2a) and ABCDEF = 682544