Puzzle for December 17, 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
eq.6 may be re-written as: A = (D + E + F) ÷ 3 Multiply both sides of the above equation by 3: 3 × A = 3 × (D + E + F) ÷ 3 which becomes eq.6a) 3×A = D + E + F
Hint #2
Add F and D to both sides of eq.2: E - F + F + D = A - D + F + F + D which becomes E + D = A + 2×F which may be written as eq.2a) D + E = A + 2×F
Hint #3
In eq.6a, replace D + E with A + 2×F (from eq.2a): 3×A = A + 2×F + F which becomes 3×A = A + 3×F Subtract A from each side of the equation above: 3×A - A = A + 3×F - A which makes 2×A = 3×F Divide both sides by 2: 2×A ÷ 2 = 3×F ÷ 2 which makes A = 1½×F
Hint #4
In eq.2a, replace A with 1½×F: D + E = 1½×F + 2×F which becomes eq.2b) D + E = 3½×F
Hint #5
In eq.4, replace A with 1½×F: 1½×F + F = C + E - 1½×F which becomes 2½×F = C + E - 1½×F Add 1½×F to both sides of the equation above: 2½×F + 1½×F = C + E - 1½×F + 1½×F which becomes eq.4a) 4×F = C + E
Hint #6
eq.5 may be written as: D + E - A = A + B - (C + E) In the above equation, substitute 3½×F for D + E (from eq.2b), 1½×F for A, and 4×F for C + E (from eq.4a): 3½×F - 1½×F = 1½×F + B - (4×F) which becomes 2×F = B - 2½×F Add 2½×F to both sides of the above equation: 2×F + 2½×F = B - 2½×F + 2½×F which makes 4½×F = B
Hint #7
Add D and C to both sides of eq.3: C - D + D + C = B - C + E + D + C which becomes 2×C = B + E + D which may be written as eq.3a) 2×C = B + D + E
Hint #8
Substitute 4½×F for B, and 3½×F for D + E (from eq.2b) in eq.3a: 2×C = 4½×F + 3½×F which becomes 2×C = 8×F Divide both sides of the above equation by 2: 2×C ÷ 2 = 8×F ÷ 2 which makes C = 4×F
Hint #9
Substitute C for 4×F in eq.4a: C = C + E Subtract C from each side of the equation above: C - C = C + E - C which means 0 = E
Hint #10
Substitute 0 for E in eq.2b: D + 0 = 3½×F which makes D = 3½×F
Solution
Substitute 1½×F for A, 4½×F for B, 4×F for C, 3½×F for D, and 0 for E in eq.1: 1½×F + 4½×F + 4×F + 3½×F + 0 + F = 29 which simplifies to 14½×F = 29 Divide both sides of the above equation by 14½: 14½×F ÷ 14½ = 29 ÷ 14½ which means F = 2 making A = 1½×F = 1½ × 2 = 3 B = 4½×F = 4½ × 2 = 9 C = 4×F = 4 × 2 = 8 D = 3½×F = 3½ × 2 = 7 and ABCDEF = 398702