Puzzle for October 20, 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
Add C to both sides of eq.5: A – C + C = C – E + C which becomes eq.5a) A = 2×C – E In eq.6, replace A with 2×C – E (from eq.5a), and D with C + E (from eq.2): F = 2×C – E + C + C + E + E which becomes eq.6a) F = 4×C + E
Hint #2
In eq.4, replace D with C + E (from eq.2), F with 4×C + E (from eq.6a), and A with 2×C – E (from eq.5a): C + E + 4×C + E = 2×C – E + B which becomes 5×C + 2×E = 2×C – E + B In the above equation, subtract 2×C from both sides, and add E to both sides: 5×C + 2×E – 2×C + E = 2×C – E + B – 2×C + E which becomes eq.4a) 3×C + 3×E = B
Hint #3
In eq.3, substitute 3×C + 3×E for B (from eq.4a), (2×C – E) for A (from eq.5a), and C + E for D (from eq.2): 3×C + 3×E – (2×C – E) = (2×C – E) + C + E which becomes 3×C + 3×E – 2×C + E = 3×C which becomes C + 4×E = 3×C Subtract C from each side of the above equation: C + 4×E – C = 3×C – C which makes 4×E = 2×C Divide both sides by 2: 4×E ÷ 2 = 2×C ÷ 2 which makes 2×E = C
Hint #4
Substitute (2×E) for C in eq.5a: A = 2×(2×E) – E which becomes A = 4×E – E which makes A = 3×E
Hint #5
Substitute (2×E) for C in eq.4a: 3×(2×E) + 3×E = B which becomes 6×E + 3×E = B which makes 9×E = B
Hint #6
Substitute 2×E for C in eq.2: D = 2×E + E which makes D = 3×E
Hint #7
Substitute (2×E) for C in eq.6a: F = 4×(2×E) + E which becomes F = 8×E + E which makes F = 9×E
Solution
Substitute 3×E for A and D, 9×E for B and F, and 2×E for C in eq.1: 3×E + 9×E + 2×E + 3×E + E + 9×E = 27 which simplifies to 27×E = 27 Divide both sides of the above equation by 27: 27×E ÷ 27 = 27 ÷ 27 which means E = 1 making A = D = 3×E = 3 × 1 = 3 B = F = 9×E = 9 × 1 = 9 C = 2×E = 2 × 1 = 2 and ABCDEF = 392319