Puzzle for December 25, 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.
* DE is a 2-digit number (not D×E).
Scratchpad
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Hint #1
eq.4 may be written as: A + D = C + E - D In the above equation, replace C + E with A + F (from eq.2): A + D = A + F - D Subtract A from both sides, and add D to both sides: A + D - A + D = A + F - D - A + D which simplifies to 2×D = F
Hint #2
eq.6 may be written as: F = (A + C + D + E) ÷ 4 Multiply both sides of the above equation by 4: 4 × F = 4 × (A + C + D + E) ÷ 4 which becomes eq.6a) 4×F = A + C + D + E
Hint #3
In eq.6a, replace F with (2×D): 4×(2×D) = A + C + D + E which becomes 8×D = A + C + D + E Subtract D from each side of the equation above: 8×D - D = A + C + D + E - D which becomes eq.6b) 7×D = A + C + E
Hint #4
Add D to both sides of eq.4: A + D + D = C - D + E + D which becomes eq.4a) A + 2×D = C + E
Hint #5
In eq.6b, substitute A + 2×D for C + E (from eq.4a): 7×D = A + A + 2×D which becomes 7×D = 2×A + 2×D Subtract 2×D from each side of the above equation: 7×D - 2×D = 2×A + 2×D - 2×D which makes 5×D = 2×A Divide both sides by 2: 5×D ÷ 2 = 2×A ÷ 2 which makes 2½×D = A
Hint #6
Substitute 2½×D for A in eq.4a: 2½×D + 2×D = C + E which becomes eq.4b) 4½×D = C + E
Hint #7
Add D and E to both sides of eq.3: B - D + D + E = D - E + D + E which becomes eq.3a) B + E = 2×D
Hint #8
Substitute B + E for 2×D (from eq.3a) into eq.4a: A + B + E = C + E Subtract E from both sides of the equation above: A + B + E - E = C + E - E which becomes eq.4c) A + B = C
Hint #9
eq.5 may be written as: 10×D + E = A + B + C + D Subtract D from both sides of the above equation: 10×D + E - D = A + B + C + D - D which becomes eq.5a) 9×D + E = A + B + C
Hint #10
Substitute C for A + B (from eq.4c) in eq.5a: 9×D + E = C + C which becomes eq.5b) 9×D + E = 2×C
Hint #11
Subtract C from each side of eq.4b: 4½×D - C = C + E - C which becomes eq.4d) 4½×D - C = E
Hint #12
Substitute 4½×D - C for E (from eq.4d) in eq.5b: 9×D + 4½×D - C = 2×C which becomes 13½×D - C = 2×C Add C to both sides of the above equation: 13½×D - C + C = 2×C + C which makes 13½×D = 3×C Divide both sides by 3: 13½×D ÷ 3 = 3×C ÷ 3 which makes 4½×D = C
Hint #13
Substitute 4½×D for C in eq.4d: 4½×D - 4½×D = E which makes 0 = E
Hint #14
Substitute 0 for E in eq.3a: B + 0 = 2×D which makes B = 2×D
Solution
Substitute 2½×D for A, 2×D for B and F, 4½×D for C, and 0 for E in eq.1: 2½×D + 2×D + 4½×D + D + 0 + 2×D = 24 which simplifies to 12×D = 24 Divide both sides of the above equation by 12: 12×D ÷ 12 = 24 ÷ 12 which means D = 2 making A = 2½×D = 2½ × 2 = 5 B = F = 2×D = 2 × 2 = 4 C = 4½×D = 4½ × 2 = 9 and ABCDEF = 549204