Puzzle for December 3, 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
In eq.2, replace F - D with E - F (from eq.3): D - C = E - F Add C and F to both sides of the above equation: D - C + C + F = E - F + C + F which becomes eq.2a) D + F = E + C
Hint #2
In eq.3, replace E - F with B - E (from eq.4): F - D = B - E Add D and E to both sides of the above equation: F - D + D + E = B - E + D + E which becomes F + E = B + D which may be written as eq.3a) B + D = F + E
Hint #3
Subtract the left and right sides of eq.2a from the left and right sides of eq.3a, respectively: B + D - (D + F) = F + E - (E + C) which becomes B + D - D - F = F + E - E - C which becomes B - F = F - C Add F and C to both sides of the equation above: B - F + F + C = F - C + F + C which becomes eq.2b) B + C = 2×F
Hint #4
In eq.6, substitute 2×F for B + C (from eq.2b): E = (2×F) ÷ F which makes E = 2
Hint #5
Substitute 2 for E in eq.3: F - D = 2 - F In the above equation, add D and F to both sides, and subtract 2 from both sides: F - D + D + F - 2 = 2 - F + D + F - 2 which becomes eq.3b) 2×F - 2 = D
Hint #6
Substitute 2 for E in eq.4: B - 2 = 2 - F Add 2 to both sides of the above equation: B - 2 + 2 = 2 - F + 2 which becomes eq.4a) B = 4 - F
Hint #7
Substitute 4 - F for B (from eq.4a) in eq.2b: 4 - F + C = 2×F In the above equation, subtract 4 from both sides, and add F to both sides: 4 - F + C - 4 + F = 2×F - 4 + F which becomes eq.2c) C = 3×F - 4
Hint #8
Substitute 3×F - 4 for C (from eq.2c), and (2×F - 2) for D (from eq.3b) in eq.1: 3×F - 4 - A = A - (2×F - 2) which becomes 3×F - 4 - A = A - 2×F + 2 In the above equation, add A and 2×F to both sides, and subtract 2 from both sides: 3×F - 4 - A + A + 2×F - 2 = A - 2×F + 2 + A + 2×F - 2 which becomes 5×F - 6 = 2×A Divide both sides by 2: (5×F - 6) ÷ 2 = 2×A ÷ 2 which becomes eq.1a) 2½×F - 3 = A
Hint #9
eq.5 may be written as: A - E = (C + D + E + F) ÷ 4 Multiply both sides of the above equation by 4: 4 × (A - E) = 4 × (C + D + E + F) ÷ 4 which becomes 4×A - 4×E = C + D + E + F Add 4×E to both sides of the above equation: 4×A - 4×E + 4×E = C + D + E + F + 4×E which becomes eq.5a) 4×A = C + D + 5×E + F
Hint #10
Substitute (2½×F - 3) for A (from eq.5b), 3×F - 4 for C (from eq.2c), 2×F - 2 for D (from eq.3b), and 2 for E in eq.5a: 4×(2½×F - 3) = 3×F - 4 + 2×F - 2 + 5×2 + F which becomes 10×F - 12 = 6×F - 6 + 10 which becomes 10×F - 12 = 6×F + 4 In the above equation, add 12 to both sides, and subtract 6×F from both sides: 10×F - 12 + 12 - 6×F = 6×F + 4 + 12 - 6×F which makes 4×F = 16 Divide both sides by 4: 4×F ÷ 4 = 16 ÷ 4 which makes F = 4
Hint #11
Substitute 4 for F in eq.1a: 2½×4 - 3 = A which becomes 10 - 3 = A which makes 7 = A
Hint #12
Substitute 4 for F in eq.4a: B = 4 - 4 which makes B = 0
Hint #13
Substitute 4 for F in eq.3b: 2×4 - 2 = D which becomes 8 - 2 = D which makes 6 = D
Solution
Substitute 4 for F in eq.2c: C = 3×4 - 4 which becomes C = 12 - 4 which makes C = 8 and makes ABCDEF = 708624