Puzzle for March 22, 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.
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Hint #1
In eq.2, replace F with A + B + E (from eq.4): B + C + D = A + B + E Subtract B from both sides of the above equation: B + C + D – B = A + B + E – B which becomes eq.2a) C + D = A + E
Hint #2
eq.3 may be written as: C = A + E + D In the equation above, replace A + E with C + D (from eq.2a): C = C + D + D which becomes C = C + 2×D Subtract C from both sides: C – C = C + 2×D – C which becomes 0 = 2×D which means 0 = D
Hint #3
In eq.2a, substitute 0 for D: C + 0 = A + E which becomes eq.2b) C = A + E In eq.5, substitute 0 for D: A + C + 0 = B which becomes eq.5a) A + C = B
Hint #4
Substitute A + E for C (from eq.2b) in eq.5a: A + A + E = B which becomes eq.5b) 2×A + E = B Substitute 2×A + E for B (from eq.5b) in eq.4: F = A + 2×A + E + E which becomes eq.4a) F = 3×A + 2×E
Hint #5
Substitute 2×A + E for B (from eq.5b), and (3×A + 2×E) for F (from eq.4a) in eq.6: 2×A + E – E = (3×A + 2×E) ÷ A which becomes 2×A = (3×A + 2×E) ÷ A Multiply both sides by A: 2×A × A = (3×A + 2×E) ÷ A × A which becomes 2×A² = 3×A + 2×E Subtract 3×A from both sides: 2×A² – 3×A = 3×A + 2×E – 3×A which becomes 2×A² – 3×A = 2×E Divide both sides by 2: (2×A² – 3×A) ÷ 2 = 2×E ÷ 2 which becomes eq.6a) A² – 1½×A = E
Hint #6
Substitute 2×A + E for B (from eq.5b), A + E for C (from eq.2b), 0 for D, and 3×A + 2×E for F (from eq.4a) in eq.1: A + 2×A + E + A + E + 0 + E + 3×A + 2×E = 19 which simplifies to 7×A + 5×E = 19 Substitute (A² – 1½×A) for E (from eq.6a) in the above equation: 7×A + 5×(A² – 1½×A) = 19 which becomes 7×A + 5×A² – 7½×A = 19 which becomes 5×A² – ½×A = 19 Subtract 19 from both sides: 5×A² – ½×A – 19 = 19 – 19 which becomes eq.1a) 5×A² – ½×A – 19 = 0
Hint #7
eq.1a is a quadratic equation in standard form. Using the quadratic equation solution formula to solve for A in eq.1a yields: A = { (–1)×(–½) ± sq.rt.[(–½)² – (4 × (5) × (–19))] } ÷ (2 × 5) which becomes A = {½ ± sq.rt.[¼ – (–380)]} ÷ 10 which becomes A = {½ ± sq.rt.(380¼)} ÷ 10 which becomes A = (½ ± 19½) ÷ 10 In the above equation, either A = (½ + 19½) ÷ 10 = 20 ÷ 10 = 2 or A = (½ – 19½) ÷ 10 = –19 ÷ 10 = –1.9 Since A must be a non-negative integer, then A ≠ –1.9 and therefore A = 2
Solution
Substitute 2 for A in eq.6a: E = 2² – 1½×2 = 4 – 3 = 1 making B = 2×A + E = 2×2 + 1 = 4 + 1 = 5 (from eq.5b) C = A + E = 2 + 1 = 3 (from eq.2b) F = 3×A + 2×E = 3×2 + 2×1 = 6 + 2 = 8 (from eq.4a) and ABCDEF = 253018