Puzzle for February 25, 2023 ( )
Scratchpad
Find the 6-digit number ABCDEF by solving the following equations:
A, B, C, D, E, and F each represent a one-digit positive integer.
Scratchpad
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Hint #1
eq.5 may be written as: A + B = C + F + D + E In the above equation, replace C + F with A + E (from eq.3): A + B = A + E + D + E which becomes A + B = A + D + 2×E Subtract A from each side: A + B - A = A + D + 2×E - A which becomes eq.5a) B = D + 2×E
Hint #2
In eq.2, replace B with D + 2×E (from eq.5a): D + 2×E = C + E Subtract 2×E from each side of the equation above: D + 2×E - 2×E = C + E - 2×E which becomes eq.2a) D = C - E
Hint #3
In eq.4, substitute (C - E) for D (from eq.2a): F - (C - E) = C + (C - E) which becomes F - C + E = 2×C - E In the equation above, add C to both sides, and subtract E from both sides: F - C + E + C - E = 2×C - E + C - E which becomes eq.4a) F = 3×C - 2×E
Hint #4
In eq.6, substitute C + E for B (from eq.2), (C - E) for D (from eq.2a), and 3×C - 2×E for F (from eq.4a): C × E = C + E - (C - E) + 3×C - 2×E which becomes C × E = C + E - C + E + 3×C - 2×E which simplifies to C × E = 3×C Divide both sides of the above equation by C: (C × E) ÷ C = (3×C) ÷ C which makes E = 3
Hint #5
Substitute 3 for E in eq.2: eq.2b) B = C + 3
Hint #6
Substitute 3 for E in eq.2a: eq.2c) D = C - 3
Hint #7
Substitute 3 for E in eq.4a: F = 3×C - 2×3 which makes eq.4b) F = 3×C - 6
Hint #8
Substitute 3×C - 6 for F (from eq.4b), and 3 for E in eq.3: C + 3×C - 6 = A + 3 which becomes 4×C - 6 = A + 3 Subtract 3 from each side of the equation above: 4×C - 6 - 3 = A + 3 - 3 which makes eq.3a) 4×C - 9 = A
Hint #9
Substitute 4×C - 9 for A (from eq.3a), C + 3 for B (from eq.2b), C - 3 for D (from eq.2c), 3 for E, and 3×C - 6 for F (from eq.4b) in eq.1: 4×C - 9 + C + 3 + C + C - 3 + 3 + 3×C - 6 = 28 which simplifies to 10×C - 12 = 28 Add 12 to both sides of the above equation: 10×C - 12 + 12 = 28 + 12 which makes 10×C = 40 Divide both sides by 10: 10×C ÷ 10 = 40 ÷ 10 which means C = 4
Solution
Since C = 4, then: A = 4×C - 9 = 4×4 - 9 = 16 - 9 = 7 (from eq.3a) B = C + 3 = 4 + 3 = 7 (from eq.2b) D = C - 3 = 4 - 3 = 1 (from eq.2c) F = 3×C - 6 = 3×4 - 6 = 12 - 6 = 6 (from eq.4b) and ABCDEF = 774136