Puzzle for May 3, 2019 ( )
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.
* BC, CD, and DE are 2-digit numbers (not B×C, C×D, or D×E).
Scratchpad
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Hint #1
Multiply both sides of eq.3 by (-1): (-1) × (F - B) = (-1) × (B - A) which becomes -F + B = -B + A which may be written as B - F = A - B In eq.5, replace A - B with B - F: B - F + F = B + E which becomes B = B + E Subtract B from each side of the equation above: B - B = B + E - B which makes 0 = E
Hint #2
In eq.4, substitute 0 for E: 0 + F = A + C which becomes eq.4a) F = A + C
Hint #3
In eq.2, substitute 0 for E, and subtract C from each side: C + D - 0 - C = A - C which becomes eq.2a) D = A - C
Hint #4
In eq.5, substitute A + C for F, and 0 for E (from eq.4a): A - B + A + C = B + 0 Add B to both sides of the above equation: A - B + A + C + B = B + 0 + B which becomes 2×A + C = 2×B Divide both sides by 2: (2×A + C) ÷ 2 = 2×B ÷ 2 which makes eq.5a) A + ½×C = B
Hint #5
eq.6 can be written as: 10×B + C - (10×C + D) = A + 10×D + E + F which is the same as 10×B + C - 10×C - D = A + 10×D + E + F Add D to both sides of the above equation: 10×B + C - 10×C - D + D = A + 10×D + E + F + D which becomes eq.6a) 10×B - 9×C = A + 11×D + E + F
Hint #6
Substitute (A + ½×C) for B (from eq.5a), (A - C) for D (from eq.2a), 0 for E, and A + C for F (from eq.4a) in eq.6a: 10×(A + ½×C) - 9×C = A + 11×(A - C) + 0 + A + C which becomes 10×A + 5×C - 9×C = A + 11×A - 11×C + A + C which becomes 10×A - 4×C = 13×A - 10×C Add (10×C - 10×A) to both sides of the above equation: 10×A - 4×C + (10×C - 10×A) = 13×A - 10×C + (10×C - 10×A) which simplifies to 6×C = 3×A Divide both sides by 3: 6×C ÷ 3 = 3×A ÷ 3 which means 2×C = A
Hint #7
Substitute 2×C for A in eq.4a: F = 2×C + C which means F = 3×C
Hint #8
Substitute 2×C for A in eq.2a: D = 2×C - C which makes D = C
Hint #9
Substitute 2×C for A in eq.5a: 2×C + ½×C = B which makes 2½×C = B
Solution
Substitute 2×C for A, 2½×C for B, C for D, 0 for E, and 3×C for F in eq.1: 2×C + 2½×C + C + C + 0 + 3×C = 19 which becomes 9½×C = 19 Divide both sides by 9½: 9½×C ÷ 9½ = 19 ÷ 9½ which means C = 2 making A = 2×C = 2 × 2 = 4 B = 2½×C = 2½ × 2 = 5 D = C = 2 F = 3×C = 3 × 2 = 6 and ABCDEF = 452206