Puzzle for February 13, 2022 ( )
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.3 may be written as: A + E = C + F + D – E In the above equation, replace C + F with D + E (from eq.2): A + E = D + E + D – E which becomes eq.3a) A + E = 2×D
Hint #2
eq.6 may be written as: B – D = (B + C + E) ÷ 3 Multiply both sides of the above equation by 3: 3 × (B – D) = 3 × (B + C + E) ÷ 3 which becomes 3×B – 3×D = B + C + E Subtract B from each side: 3×B – 3×D – B = B + C + E – B which becomes eq.6a) 2×B – 3×D = C + E
Hint #3
In eq.4, add B to both sides, and subtract E from both sides: B + E + B – E = A – B + C + D + B – E which becomes 2×B = A + C + D – E In eq.6a, replace 2×B with A + C + D – E: A + C + D – E – 3×D = C + E which becomes A + C – 2×D – E = C + E In the above equation, subtract C from both sides, and add E to both sides: A + C – 2×D – E – C + E = C + E – C + E which becomes eq.4a) A – 2×D = 2×E
Hint #4
In eq.4a, substitute (A + E) for 2×D (from eq.3a): A – (A + E) = 2×E which becomes A – A – E = 2×E which becomes –E = 2×E Add E to both sides of the above equation: –E + E = 2×E + E which makes 0 = 3×E which means 0 = E
Hint #5
Substitute 0 for E in eq.3a: A + 0 = 2×D which makes eq.3b) A = 2×D
Hint #6
Substitute 0 for E in eq.2: C + F = D + 0 which becomes eq.2a) C + F = D
Hint #7
Substitute (C + F) for D (from eq.2a), and 0 for E in eq.6a: 2×B – 3×(C + F) = C + 0 which becomes 2×B – 3×C – 3×F = C Add 3×C and 3×F to both sides of the above equation: 2×B – 3×C – 3×F + 3×C + 3×F = C + 3×C + 3×F which becomes eq.6b) 2×B = 4×C + 3×F
Hint #8
Substitute (C + F) for D (from eq.2a) in eq.3b: A = 2×(C + F) which becomes eq.3c) A = 2×C + 2×F
Hint #9
Substitute (2×C + 2×F) for A (from eq.3c), and (C + F) for D (from eq.2a) in eq.5: (2×C + 2×F) – B – (C + F) = B – (2×C + 2×F) – F which becomes 2×C + 2×F – B – C – F = B – 2×C – 2×F – F which becomes C + F – B = B – 2×C – 3×F Add B, 2×C, and 3×F to both sides of the equation above: C + F – B + B + 2×C + 3×F = B – 2×C – 3×F + B + 2×C + 3×F which becomes eq.5a) 3×C + 4×F = 2×B
Hint #10
Substitute 3×C + 4×F for 2×B (from eq.5a) into eq.6b: 3×C + 4×F = 4×C + 3×F Subtract 3×C and 3×F from each side of the above equation: 3×C + 4×F – 3×C – 3×F = 4×C + 3×F – 3×C – 3×F which simplifies to F = C
Hint #11
Substitute C for F in eq.2a: C + C = D which makes 2×C = D
Hint #12
Substitute (2×C) for D in eq.3b: A = 2×(2×C) which makes A = 4×C
Hint #13
Substitute C for F in eq.6b: 2×B = 4×C + 3×C which makes 2×B = 7×C Divide both sides of the above equation by 2: 2×B ÷ 2 = 7×C ÷ 2 which makes B = 3½×C
Solution
Substitute 4×C for A, 3½×C for B, 2×C for D, 0 for E, and C for F in eq.1: 4×C + 3½×C + C + 2×C + 0 + C = 23 which simplifies to 11½×C = 23 Divide both sides of the above equation by 11½: 11½×C ÷ 11½ = 23 ÷ 11½ which means C = 2 making A = 4×C = 4 × 2 = 8 B = 3½×C = 3½ × 2 = 7 D = 2×C = 2 × 2 = 4 F = C = 2 and ABCDEF = 872402