Puzzle for April 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 D to both sides of eq.4: D + E + D = A + C - D + D which becomes eq.4a) 2×D + E = A + C Add A and D to both sides of eq.5: D - A + A + D = C - D + E + A + D which becomes 2×D = C + E + A which may be written as eq.5a) 2×D = A + C + E
Hint #2
In eq.5a, replace A + C with 2×D + E (from eq.4a): 2×D = 2×D + E + E which becomes 2×D = 2×D + 2×E Subtract 2×D from each side of the equation above: 2×D - 2×D = 2×D + 2×E - 2×D which makes 0 = 2×E which means 0 = E
Hint #3
In eq.3, substitute 0 for E: A - 0 = D - A which becomes A = D - A Add A to both sides of the above equation: A + A = D - A + A which makes 2×A = D
Hint #4
In eq.5a, substitute (2×A) for D, and 0 for E: 2×(2×A) = A + C + 0 which becomes 4×A = A + C Subtract A from both sides of the equation above: 4×A - A = A + C - A which makes 3×A = C
Hint #5
In eq.2, substitute 0 for E, and 2×A for D: B + 0 = A + 2×A which makes B = 3×A
Hint #6
Subtract F from both sides of eq.6: C + D - F - F = average (A, B, C) + F - F which becomes eq.6a) C + D - 2×F = average (A, B, C)
Hint #7
eq.6a may be re-written as: C + D - 2×F = (A + B + C) ÷ 3 Multiply both sides of the above equation by 3: 3 × (C + D - 2×F) = 3 × (A + B + C) ÷ 3 which becomes 3×C + 3×D - 6×F = A + B + C Subtract C from both sides: 3×C + 3×D - 6×F - C = A + B + C - C which becomes eq.6b) 2×C + 3×D - 6×F = A + B
Hint #8
Substitute (3×A) for C and B, and (2×A) for D in eq.6b: 2×(3×A) + 3×(2×A) - 6×F = A + (3×A) which becomes 6×A + 6×A - 6×F = 4×A which becomes 12×A - 6×F = 4×A In the equation above, add 6×F to both sides, and subtract 4×A from both sides: 12×A - 6×F + 6×F - 4×A = 4×A + 6×F - 4×A which becomes 8×A = 6×F Divide both sides by 6: 8×A ÷ 6 = 6×F ÷ 6 which makes 1⅓×A = F
Solution
Substitute 3×A for B and C, 2×A for D, 0 for E, and 1⅓×A for F in eq.1: A + 3×A + 3×A + 2×A + 0 + 1⅓×A = 31 which simplifies to 10⅓×A = 31 Divide both sides of the above equation by 10⅓: 10⅓×A ÷ 10⅓ = 31 ÷ 10⅓ which means A = 3 making B = C = 3×A = 3 × 3 = 9 D = 2×A = 2 × 3 = 6 F = 1⅓×A = 1⅓ × 3 = 4 and ABCDEF = 399604