Puzzle for October 27, 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.4, replace E with A + D (from eq.2): A + F = B + A + D Subtract A from each side of the equation above: A + F - A = B + A + D - A which becomes eq.4a) F = B + D
Hint #2
In eq.5, substitute (B + D) for F (from eq.4a): D - B = B - (B + D) which becomes D - B = B - B - D which becomes D - B = -D Add B and D to both sides of the above equation: D - B + B + D = -D + B + D which makes 2×D = B
Hint #3
In eq.5, replace B with 2×D (from eq.4a): D - 2×D = 2×D - F which makes -D = 2×D - F Add D and F to both sides of the equation above: -D + D + F = 2×D - F + D + F which makes F = 3×D
Hint #4
Substitute 2×D for B in eq.3: C = E - 2×D Add 2×D to both sides of the above equation: C + 2×D = E - 2×D + 2×D which becomes eq.3a) C + 2×D = E
Hint #5
Substitute 2×D for B, 3×D for F, and A + D for E (from eq.2) into eq.6: 2×D + 3×D = A + C + D + A + D which becomes 5×D = 2×A + C + 2×D Subtract 2×A and 2×D from each side of the above equation: 5×D - 2×A - 2×D = 2×A + C + 2×D - 2×A - 2×D which becomes eq.6a) 3×D - 2×A = C
Hint #6
Substitute 3×D for F, 2×D for B, and C + 2×D for E (from eq.3a) in eq.4: A + 3×D = 2×D + C + 2×D which becomes A + 3×D = 4×D + C Subract 4×D from both sides of the equation above: A + 3×D - 4×D = 4×D + C - 4×D which becomes eq.4b) A - D = C
Hint #7
Substitute 3×D - 2×A for C (from eq.6a) in eq.4b: A - D = 3×D - 2×A Add D and 2×A to both sides of the above equation: A - D + D + 2×A = 3×D - 2×A + D + 2×A which makes 3×A = 4×D Divide both sides by 3: 3×A ÷ 3 = 4×D ÷ 3 which makes A = 1⅓×D
Hint #8
Substitute 1⅓×D for A in eq.4b: 1⅓×D - D = C which makes ⅓×D = C
Hint #9
Substitute ⅓×D for C in eq.3a: ⅓×D + 2×D = E which makes 2⅓×D = E
Solution
Substitute 1⅓×D for A, 2×D for B, ⅓×D for C, 2⅓×D for E, and 3×D for F in eq.1: 1⅓×D + 2×D + ⅓×D + D + 2⅓×D + 3×D = 30 which simplifies to 10×D = 30 Divide both sides of the above equation by 10: 10×D ÷ 10 = 30 ÷ 10 which means D = 3 making A = 1⅓×D = 1⅓ × 3 = 4 B = 2×D = 2 × 3 = 6 C = ⅓×D = ⅓ × 3 = 1 E = 2⅓×D = 2⅓ × 3 = 7 F = 3×D = 3 × 3 = 9 and ABCDEF = 461379