Puzzle for March 20, 2023 ( )
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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.
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Hint #1
In eq.5, substitute (A + F) for D (from eq.2): F - (A + F) = A - F which becomes F - A - F = A - F which becomes -A = A - F Add A and F to both sides of the above equation: -A + A + F = A - F + A + F which makes F = 2×A
Hint #2
In eq.2, replace F with 2×A: D = A + 2×A which makes D = 3×A
Hint #3
In eq.6, replace D with 3×A, and F with 2×A: C - 3×A = 2×A - C Add 3×A and C to both sides of the equation above: C - 3×A + 3×A + C = 2×A - C + 3×A + C which makes 2×C = 5×A Divide both sides by 2: 2×C ÷ 2 = 5×A ÷ 2 which makes C = 2½×A
Hint #4
In eq.3, substitute 2½×A for C: E = A + 2½×A which makes E = 3½×A
Hint #5
Substitute 3½×A for E in eq.4: B = A + 3½×A which makes B = 4½×A
Solution
Substitute 4½×A for B, 2½×A for C, 3×A for D, 3½×A for E, and 2×A for F in eq.1: A + 4½×A + 2½×A + 3×A + 3½×A + 2×A = 33 which simplifies to 16½×A = 33 Divide both sides of the above equation by 16½: 16½×A ÷ 16½ = 33 ÷ 16½ which means A = 2 making B = 4½×A = 4½ × 2 = 9 C = 2½×A = 2½ × 2 = 5 D = 3×A = 3 × 2 = 6 E = 3½×A = 3½ × 2 = 7 F = 2×A = 2 × 2 = 4 and ABCDEF = 295674