Puzzle for May 21, 2020 ( )
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.6: C – D + D = B + F + D which becomes C = B + F + D In eq.4, replace C with B + F + D: B + F + D + D = E + F which becomes B + F + 2×D = E + F Subtract F from both sides of the equation above: B + F + 2×D – F = E + F – F which becomes eq.4a) B + 2×D = E
Hint #2
In eq.2, replace E with B + 2×D (from eq.4a): D + B + 2×D = A + B which becomes B + 3×D = A + B Subtract B from both sides of the above equation: B + 3×D – B = A + B – B which makes 3×D = A
Hint #3
In eq.6, substitute A – B for F (from eq.5): C – D = B + A – B which becomes eq.6a) C – D = A
Hint #4
Substitute 3×D for A in eq.6a: C – D = 3×D Add D to both sides of the equation above: C – D + D = 3×D + D which makes C = 4×D
Hint #5
Substitute B + 2×D for E (from eq.4a), 3×D for A, and 4×D for C in eq.3: B + B + 2×D = 3×D + 4×D which becomes 2×B + 2×D = 7×D Subtract 2×D from each side of the above equation above: 2×B + 2×D – 2×D = 7×D – 2×D which makes 2×B = 5×D Divide both sides by 2: 2×B ÷ 2 = 5×D ÷ 2 which means B = 2½×D
Hint #6
Substitute 2½×D for B in eq.4a: 2½×D + 2×D = E which makes 4½×D = E
Hint #7
Substitute 3×D for A, and 2½×D for B in eq.5: F = 3×D – 2½×D which makes F = ½×D
Solution
Substitute 3×D for A, 2½×D for B, 4×D for C, 4½×D for E, and ½×D for F in eq.1: 3×D + 2½×D + 4×D + D + 4½×D + ½×D = 31 which simplifies to 15½×D = 31 Divide both sides of the equation above by 15½: 15½×D ÷ 15½ = 31 ÷ 15½ which means D = 2 making A = 3×D = 3 × 2 = 6 B = 2½×D = 2½ × 2 = 5 C = 4×D = 4 × 2 = 8 E = 4½×D = 4½ × 2 = 9 F = ½×D = ½ × 2 = 1 and ABCDEF = 658291