Puzzle for May 23, 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 E and A to both sides of eq.2: C – E + E + A = D – A + E + A which becomes eq.2a) C + A = D + E In eq.4, replace D + E with C + A (from eq.2a): B + C = C + A Subtract C from each side of the equation above: B + C – C = C + A – C which makes B = A
Hint #2
In eq.3, replace B with A: D = A – A which makes D = 0
Hint #3
In eq.2a, substitute 0 for D: C + A = 0 + E which becomes eq.2b) C + A = E
Hint #4
Substitute A for B, 0 for D, and C + A for E (from eq.2b) in eq.1: A + A – 0 + C + A – F = C + 0 + F which becomes 3×A + C – F = C + F In the above equation, subtract C from both sides, and add F to both sides: 3×A + C – F – C + F = C + F – C + F which makes 3×A = 2×F Divide both sides by 2: 3×A ÷ 2 = 2×F ÷ 2 which makes 1½×A = F
Hint #5
Substitute A for B, and 1½×A for F in eq.6: C = ((A ÷ C) + 1½×A) ÷ A Multiply both sides of the equation above by A: A × C = A × ((A ÷ C) + 1½×A) ÷ A which becomes A × C = (A ÷ C) + 1½×A Subtract (A ÷ C) from both sides: A × C – (A ÷ C) = (A ÷ C) + 1½×A – (A ÷ C) which may be written as eq.6a) A × (C – (1 ÷ C)) = 1½×A
Hint #6
Divide both sides of eq.6a by A: A × (C – (1 ÷ C)) ÷ A = 1½×A ÷ A which becomes C – (1 ÷ C) = 1½ Multiply both sides of the equation above by C: C – (1 ÷ C) × C = 1½ × C which becomes C² – 1 = 1½×C Subtract 1½×C from each side: C² – 1 – 1½×C = 1½×C – 1½×C which is the same as eq.6b) C² – 1½×C – 1 = 0
Hint #7
eq.6b is a quadratic equation in standard form. Using the quadratic equation solution formula to solve for C in eq.6b yields: C = { (–1)×(–1½) ± sq.rt.[(–1½)² – (4 × 1 × (–1))] } ÷ (2 × 1) which becomes C = {1½ ± sq.rt.(2¼ – (–4))} ÷ 2 which becomes C = (1½ ± sq.rt.(6¼)) ÷ 2 which becomes C = (1½ ± 2½) ÷ 2 In the above equation, either C = (1½ + 2½) ÷ 2 = 4 ÷ 2 = 2 or C = (1½ – 2½) ÷ 2 = –1 ÷ 2 = –½ Since C must be a non-negative integer, then C ≠ –½ and therefore means C = 2
Hint #8
Substitute 2 for C in eq.2b: eq.2c) 2 + A = E
Solution
Substitute (2 + A) for E (from eq.2c), 2 for C, and A for B in eq.5: (2 + A) ÷ 2 = A – 2 Multiply both sides of the equation above by 2: 2 × (2 + A) ÷ 2 = 2 × (A – 2) which becomes 2 + A = 2×A – 4 Subtract A from both sides, and add 4 to both sides: 2 + A – A + 4 = 2×A – 4 – A + 4 which makes 6 = A and also makes B = A = 6 and makes E = 2 + A = 2 + 6 = 8 (from eq.2c) F = 1½×A = 1½ × 6 = 9 and ABCDEF = 662089