Puzzle for July 2, 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
In eq.3, replace A with C + F (from eq.2): D + E + F = C + F Subtract F from each side of the equation above: D + E + F – F = C + F – F which becomes eq.3a) D + E = C
Hint #2
In eq.4, replace C with D + E (from eq.3a): E + F = D + E + D which becomes E + F = 2×D + E Subtract E from both sides of the above equation: E + F – E = 2×D + E – E which makes F = 2×D
Hint #3
In eq.2, replace C with D + E (from eq.3a), and F with 2×D: D + E + 2×D = A which makes eq.2a) 3×D + E = A
Hint #4
In eq.5, substitute D + E for C (from eq.3a), 3×D + E for A (from eq.2a), and 2×D for F: B + D + E = 3×D + E + E + 2×D which becomes B + D + E = 5×D + 2×E Subtract D and E from both sides of the equation above: B + D + E – D – E = 5×D + 2×E – D – E which becomes eq.5a) B = 4×D + E
Hint #5
eq.6 may be written as: C = (B + D + F) ÷ 3 Multiply both sides of the above equation by 3: 3 × C = 3 × (B + D + F) ÷ 3 which becomes eq.6a) 3×C = B + D + F
Hint #6
Substitute (D + E) for C (from eq.3a), 4×D + E for B (from eq.5a), and 2×D for F in eq.6a: 3×(D + E) = 4×D + E + D + 2×D which becomes 3×D + 3×E = 7×D + E Subtract 3×D and E from each side of the equation above: 3×D + 3×E – 3×D – E = 7×D + E – 3×D – E which makes 2×E = 4×D Divide both sides by 2: 2×E ÷ 2 = 4×D ÷ 2 which makes E = 2×D
Hint #7
Substitute 2×D for E in eq.2a: 3×D + 2×D = A which makes 5×D = A
Hint #8
Substitute 2×D for E in eq.5a: B = 4×D + 2×D which makes B = 6×D
Hint #9
Substitute 2×D for E in eq.3a: D + 2×D = C which makes 3×D = C
Solution
Substitute 5×D for A, 6×D for B, 3×D for C, and 2×D for E and F in eq.1: 5×D + 6×D + 3×D + D + 2×D + 2×D = 19 which simplifies to 19×D = 19 Divide both sides of the above equation by 19: 19×D ÷ 19 = 19 ÷ 19 which means D = 1 making A = 5×D = 5 × 1 = 5 B = 6×D = 6 × 1 = 6 C = 3×D = 3 × 1 = 3 E = F = 2×D = 2 × 1 = 2 and ABCDEF = 563122