Puzzle for September 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
In eq.4, add C to both sides, and subtract D from both sides: D + E + C - D = A - C + F + C - D which becomes E + C = A + F - D which may be written as eq.4a) C + E = A + F - D
Hint #2
In eq.3, replace C + E with A + F - D (from eq.4a): A + F - D = B + F In the above equation, subtract F from both sides, and add D to both sides: A + F - D - F + D = B + F - F + D which becomes eq.3a) A = B + D
Hint #3
In eq.2, replace A with B + D (from eq.2): B - C = B + D - D which becomes B - C = B Subtract B from each side of the equation above: B - C - B = B - B which makes -C = 0 which means C = 0
Hint #4
In eq.3, substitute 0 for C: 0 + E = B + F which becomes eq.3b) E = B + F
Hint #5
Substitute B + F for E (from eq.3b), (B + D) for A (from eq.3a), and 0 for C in eq.5: B + F - (B + D) = (B + D) - B + 0 which becomes B + F - B - D = D which becomes F - D = D Add D to both sides of the above equation: F - D + D = D + D which becomes eq.5a) F = 2×D
Hint #6
eq.6 may be written as: F - B = (A + C + D + E) ÷ 4 Multiply both sides of the above equation by 4: 4 × (F - B) = 4 × (A + C + D + E) ÷ 4 which becomes eq.6a) 4×F - 4×B = A + C + D + E
Hint #7
Substitute 0 for C, B + D for A (from eq.3a), and B + F for E (from eq.3b) in eq.6a: 4×F - 4×B = B + D + 0 + D + B + F which becomes 4×F - 4×B = 2×B + 2×D + F Subtract 2×B and F from both sides of the equation above: 4×F - 4×B - 2×B - F = 2×B + 2×D + F - 2×B - F which becomes eq.6b) 3×F - 6×B = 2×D
Hint #8
Substitute F for 2×D (from eq.5a) in eq.6b: 3×F - 6×B = F In the equation above, add 6×B to both sides, and subtract F from both sides: 3×F - 6×B + 6×B - F = F + 6×B - F which becomes 2×F = 6×B Divide both sides by 2: 2×F ÷ 2 = 6×B ÷ 2 which makes F = 3×B
Hint #9
Substitute 3×B for F in eq.3b: E = B + 3×B which makes E = 4×B
Hint #10
Substitute 3×B for F in eq.5a: 3×B = 2×D Divide both sides of the above equation by 2: 3×B ÷ 2 = 2×D ÷ 2 which makes 1½×B = D
Hint #11
Substitute 1½×B for D in eq.3a: A = B + 1½×B which makes A = 2½×B
Solution
Substitute 2½×B for A, 0 for C, 1½×B for D, 4×B for E, and 3×B for F in eq.1: 2½×B + B + 0 + 1½×B + 4×B + 3×B = 24 which simplifies to 12×B = 24 Divide both sides of the above equation by 12: 12×B ÷ 12 = 24 ÷ 12 which means B = 2 making A = 2½×B = 2½ × 2 = 5 D = 1½×B = 1½ × 2 = 3 E = 4×B = 4 × 2 = 8 F = 3×B = 3 × 2 = 6 and ABCDEF = 520386