Puzzle for November 14, 2021 ( )
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 B to both sides of eq.3: A + B – D + B = C + E – B + B which becomes eq.3a) A + 2×B – D = C + E In eq.2, replace F with B + D (from eq.1): C + E = A + B + B + D which becomes eq.2a) C + E = A + 2×B + D
Hint #2
In eq.2a, replace C + E with A + 2×B – D (from eq.3a): A + 2×B – D = A + 2×B + D Subtract A, 2×B, and D from both sides of the equation above: A + 2×B – D – A – 2×B – D = A + 2×B + D – A – 2×B – D which simplifies to –2×D = 0 which means D = 0
Hint #3
In eq.1, substitute 0 for D: B + 0 = F which makes B = F
Hint #4
Substitute 0 for D, and B for F in eq.6: A = (A + C – 0 + E + B) ÷ B which becomes A = (A + C + E + B) ÷ B Multiply both sides of the above equation by B: A × B = ((A + C + E + B) ÷ B) × B which becomes eq.6a) A × B = A + C + E + B
Hint #5
Substitute 0 for D, and B for F in eq.4: C = (A + C + 0 + E) ÷ B which becomes C = (A + C + E) ÷ B Multiply both sides of the above equation by B: C × B = (A + C + E) ÷ B) × B which becomes eq.4a) C × B = A + C + E
Hint #6
Substitute (C × B) for A + C + E (from eq.4a) into eq.6a: A × B = (C × B) + B which is equivalent to A × B = (C + 1) × B Since B ≠ 0 (from eq.6), divide both sides of the above equation by B: (A × B) ÷ B = ((C + 1) × B) ÷ B which makes eq.6b) A = C + 1
Hint #7
Substitute C + 1 for A (from eq.6b), and B for F in eq.2: C + E = C + 1 + B + B which becomes C + E = C + 1 + 2×B Subtract C from both sides of the above equation: C + E – C = C + 1 + 2×B – C which becomes eq.2b) E = 1 + 2×B
Hint #8
eq.5 may be written as: E = (A + C + E – B – D) ÷ F Substitute (C × B) for A + C + E (from eq.4a), 0 for D, and B for F in the above equation: E = ((C × B) – B – 0) ÷ B which becomes E = ((C × B) – B) ÷ B which is equivalent to eq.5a) E = C – 1
Hint #9
Substitute C – 1 for E (from eq.5a) in eq.2b: C – 1 = 1 + 2×B Add 1 to both sides of the equation above: C – 1 + 1 = 1 + 2×B + 1 which becomes eq.2c) C = 2 + 2×B
Hint #10
Substitute 2 + 2×B for C (from eq.2c) in eq.6b: A = 2 + 2×B + 1 which becomes eq.6c) A = 3 + 2×B
Hint #11
Substitute (2 + 2×B) for C (from eq.2b), (3 + 2×B) for A (from eq.6c), and (1 + 2×B) for E (from eq.2a) in eq.4a: (2 + 2×B) × B = (3 + 2×B) + (2 + 2×B) + (1 + 2×B) which becomes 2×B + 2×B² = 6 + 6×B Subtract 6 and 6×B from each side of the above equation: 2×B + 2×B² – 6 – 6×B = 6 + 6×B – 6 – 6×B which becomes 2×B² – 6 – 4×B = 0 Divide both sides of the above equation by 2: (2×B² – 6 – 4×B) ÷ 2 = 0 ÷ 2 which becomes B² – 3 – 2×B = 0 which may be written as: eq.6d) B² – 2×B – 3 = 0
Solution
eq.6d is a quadratic equation in standard form. Using the quadratic equation solution formula to solve for B in eq.6d yields: B = { (–1)×(–2) ± sq.rt.[(–2)² – (4 × (1) × (–3))] } ÷ (2 × (1)) which becomes B = {2 ± sq.rt.(4 – (–12))} ÷ 2 which becomes B = {2 ± sq.rt.(16)} ÷ 2 which becomes B = (2 ± 4) ÷ 2 In the above equation, either B = (2 + 4) ÷ 2 = 6 ÷ 2 = 3 or B = (2 – 4) ÷ 2 = –2 ÷ 2 = –1 Since B must be a non-negative integer, then B ≠ –1 and therefore makes B = 3 and makes A = 3 + 2×B = 3 + 2×3 = 3 + 6 = 9 (from eq.6c) C = 2 + 2×B = 2 + 2×3 = 2 + 6 = 8 (from eq.2c) E = 1 + 2×B = 1 + 2×3 = 1 + 6 = 7 (from eq.2b) F = B = 3 and ABCDEF = 938073