Puzzle for July 25, 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 F to both sides of eq.4: B + E + F = C – F + F which becomes B + E + F = C In the above equation, replace E + F with B + D (from eq.2): B + B + D = C which becomes eq.4a) 2×B + D = C
Hint #2
In eq.3, replace C with 2×B + D (from eq.4a): D + F = B + 2×B + D which becomes D + F = 3×B + D Subtract D from each side of the above equation: D + F – D = 3×B + D – D which makes F = 3×B
Hint #3
In eq.2, substitute 3×B for F: E + 3×B = B + D Subtract B and E from each side of the equation above: E + 3×B – B – E = B + D – B – E which becomes eq.2a) 2×B = D – E
Hint #4
Since B is non-negative, then in eq.2a: B ≥ 0 which means ie.1a) 2×B ≥ 0 In the inequality ie.1a, substitute D – E for 2×B (from eq.2a): D – E ≥ 0 Add E to both sides of the above inequality: D – E + E ≥ 0 + E which means ie.1b) D ≥ E
Hint #5
Substitute 3×B for F in eq.3: D + 3×B = B + C Subtract D and B from both sides of the above equation: D + 3×B – D – B = B + C – D – B which becomes eq.3a) 2×B = C – D
Hint #6
Substitute C – D for 2×B (from eq.3a) in inequality ie.1a: ie.2a) C – D ≥ 0 Add D to both sides of inequality ie.2a: C – D + D ≥ 0 + D which means ie.2b) C ≥ D
Hint #7
eq.6 may be written as: C = (D + E) ÷ 2 Multiply both sides of the equation above by 2: C × 2 = ((D + E) ÷ 2) × 2 which becomes 2×C = D + E Subtract C and D from both sides: 2×C – C – D = D + E – C – D which becomes eq.6a) C – D = E – C
Hint #8
Substitute E – C for C – D (from eq.6a) in inequality ie.2a: ie.6a) E – C ≥ 0 Add C to both sides of inequlity ie.6a: E – C + C ≥ 0 + C which means ie.6b) E ≥ C
Hint #9
Combine C ≥ D (from ie.2b), D ≥ E (from ie.1b), and E ≥ C (from ie.6b): C ≥ D ≥ E ≥ C The only values for C, D, and E that make the above inequality true are: C = D = E
Hint #10
Substitute C for D in eq.3a: 2×B = C – C which makes 2×B = 0 which means B = 0 and also makes F = 3×B = 3 × 0 = 0
Hint #11
Substitute C for D and E in eq.5: A – (C ÷ C) = C + C + C – A which becomes A – 1 = 3×C – A Add 1 and A to both sides of the above equation: A – 1 + 1 + A = 3×C – A + 1 + A which becomes 2×A = 3×C + 1 Divide both sides by 2: 2×A ÷ 2 = (3×C + 1) ÷ 2 which becomes eq.5a) A = 1½×C + ½
Solution
Substitute 1½×C + ½ for A, 0 for B and F, and C for D and E in eq.1: 1½×C + ½ + 0 + C + C + C + 0 = 23 which simplifies to 4½×C + ½ = 23 Subtract ½ from both sides of the above equation: 4½×C + ½ – ½ = 23 – ½ which means 4½×C = 22½ Divide both sides by 4½: 4½×C ÷ 4½ = 22½ ÷ 4½ which makes C = 5 making D = E = C = 5 A = 1½×C + ½ = (1½ × 5) + ½ = 7½ + ½ = 8 (from eq.5a) and ABCDEF = 805550