Puzzle for November 26, 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.
* EF and DE are 2-digit numbers (not E×F or D×E).
Scratchpad
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Hint #1
eq.6 may be written as: D + F = (C + E) ÷ 2 Multiply both sides of the above equation by 2: 2 × (D + F) = 2 × (C + E) ÷ 2 which becomes eq.6a) 2×D + 2×F = C + E
Hint #2
Add the left and right sides of eq.6a to the left and right sides of eq.3, respectively: B + D + E - F + 2×D + 2×F = A + F + C + E which becomes B + 3×D + E + F = A + F + C + E Subtract B, E, and F from each side of the equation above: B + 3×D + E + F - B - E - F = A + F + C + E - B - E - F which simplifies to eq.3a) 3×D = A + C - B
Hint #3
eq.5 may be written as: D = (A + B + C) ÷ 3 Multiply both sides of the above equation by 3: 3 × D = 3 × (A + B + C) ÷ 3 which becomes eq.5a) 3×D = A + B + C
Hint #4
In eq.5a, replace 3×D with A + C - B (from eq.3a): A + C - B = A + B + C Subtract A and C from both sides of the above equation: A + C - B - A - C = A + B + C - A - C which simplifies to -B = B Add B to both sides: -B + B = B + B which makes 0 = 2×B which means 0 = B
Hint #5
In eq.2, replace B with 0: E + F = A + 0 which becomes eq.2a) E + F = A
Hint #6
In eq.3, substitute 0 for B, and E + F for A (from eq.2a): 0 + D + E - F = E + F + F which becomes D + E - F = E + 2×F In the equation above, subtract E from both sides, and add F to both sides: D + E - F - E + F = E + 2×F - E + F which becomes D = 3×F
Hint #7
eq.4 may be written as: 10×E + F - (10×D + E) = E - F which becomes 10×E + F - 10×D - E = E - F which becomes 9×E + F - 10×D = E - F In the above equation, add 10×D and F to both sides, and subtract E from both sides: 9×E + F - 10×D + 10×D + F - E = E - F + 10×D + F - E which becomes eq.4a) 8×E + 2×F = 10×D
Hint #8
Substitute (3×F) for D in eq.4a: 8×E + 2×F = 10×(3×F) which becomes 8×E + 2×F = 30×F Subtract 2×F from each side of the equation above: 8×E + 2×F - 2×F = 30×F - 2×F which makes 8×E = 28×F Divide both sides by 8: 8×E ÷ 8 = 28×F ÷ 8 which makes E = 3½×F
Hint #9
Substitute 3½×F for E in eq.2a: 3½×F + F = A which makes 4½×F = A
Hint #10
Substitute (3×F) for D, and 3½×F for E in eq.6a: 2×(3×F) + 2×F = C + 3½×F which becomes 6×F + 2×F = C + 3½×F which becomes 8×F = C + 3½×F Subtract 3½×F from each side of the above equation: 8×F - 3½×F = C + 3½×F - 3½×F which makes 4½×F = C
Solution
Substitute 4½×F for A and C, 0 for B, 3×F for D, and 3½×F for E in eq.1: 4½×F + 0 + 4½×F + 3×F + 3½×F + F = 33 which simplifies to 16½×F = 33 Divide both sides of the above equation by 16½: 16½×F ÷ 16½ = 33 ÷ 16½ which means F = 2 making A = C = 4½×F = 4½ × 2 = 9 D = 3×F = 3 × 2 = 6 E = 3½×F = 3½ × 2 = 7 and ABCDEF = 909672