Puzzle for March 12, 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 non-negative integer.
Scratchpad
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Hint #1
Add A and F to both sides of eq.3: B - A + A + F = C - F + A + F which becomes B + F = C + A which may be written as eq.3a) B + F = A + C
Hint #2
In eq.4, replace B + F with A + C (from eq.3a): C + D - A = A + A + C which becomes C + D - A = 2×A + C In the above equation, subtract C from both sides, and add A to both sides: C + D - A - C + A = 2×A + C - C + A which makes D = 3×A
Hint #3
Add E to both sides of eq.5: E + F + E = A + B + C + D - E + E which becomes 2×E + F = A + B + C + D which may be written as eq.5a) 2×E + F = A + C + B + D
Hint #4
In eq.5a, replace A + C with B + F (from eq.3a), and D with 3×A: 2×E + F = B + F + B + 3×A which becomes 2×E + F = 2×B + F + 3×A Subtract F from both sides of the above equation: 2×E + F - F = 2×B + F + 3×A - F which becomes eq.5b) 2×E = 2×B + 3×A
Hint #5
eq.6 may be written as: E - A - F = (B + C + D + F) ÷ 4 Multiply both sides of the above equation by 4: 4 × (E - A - F) = 4 × (B + C + D + F) ÷ 4 which becomes 4×E - 4×A - 4×F = B + C + D + F Add 4×A and 4×F to both sides: 4×E - 4×A - 4×F + 4×A + 4×F = B + C + D + F + 4×A + 4×F which becomes 4×E = B + C + D + 5×F + 4×A which may be written as eq.6a) 4×E = A + C + B + D + 5×F + 3×A
Hint #6
In eq.6a, substitute 2×E + F for A + C + B + D (from eq.5a): 4×E = 2×E + F + 5×F + 3×A which becomes 4×E = 2×E + 6×F + 3×A Subtract 2×E from each side of the equation above: 4×E - 2×E = 2×E + 6×F + 3×A - 2×E which becomes eq.6b) 2×E = 6×F + 3×A
Hint #7
In eq.5b, substitute 6×F + 3×A for 2×E (from eq.6b): 6×F + 3×A = 2×B + 3×A Subtract 3×A from each side of the above equation: 6×F + 3×A - 3×A = 2×B + 3×A - 3×A which makes 6×F = 2×B Divide both sides by 2: 6×F ÷ 2 = 2×B ÷ 2 which makes eq.5c) 3×F = B
Hint #8
Substitute 3×A for D in eq.2: A + 3×A = C + F which becomes eq.2a) 4×A = C + F
Hint #9
Substitute 3×F for B in eq.3: 3×F - A = C - F Add F to both sides of the above equation: 3×F - A + F = C - F + F which becomes eq.3b) 4×F - A = C
Hint #10
Substitute 4×F - A for C (from eq.3b) into eq.2a: 4×A = 4×F - A + F which becomes 4×A = 5×F - A Add A to both sides of the above equation: 4×A + A = 5×F - A + A which makes 5×A = 5×F Divide both sides by 5: 5×A ÷ 5 = 5×F ÷ 5 which makes A = F
Hint #11
Substitute A for F in eq.5c: 3×A = B
Hint #12
Substitute A for F in eq.3b: 4×A - A = C which makes 3×A = C
Hint #13
Substitute (3×A) for B in eq.5b: 2×E = 2×(3×A) + 3×A which becomes 2×E = 6×A + 3×A which makes 2×E = 9×A Divide both sides of the above equation by 2: 2×E ÷ 2 = 9×A ÷ 2 which makes E = 4½×A
Solution
Substitute 3×A for B and C and D, 4½×A for E, and A for F in eq.1: A + 3×A + 3×A + 3×A + 4½×A + A = 31 which simplifies to 15½×A = 31 Divide both sides of the above equation by 15½: 15½×A ÷ 15½ = 31 ÷ 15½ which means A = 2 making B = C = D = 3×A = 3 × 2 = 6 E = 4½×A = 4½ × 2 = 9 F = A = 2 and ABCDEF = 266692