Puzzle for November 6, 2022 ( )
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 B and A to both sides of eq.2: E - B + B + A = C - A + B + A which becomes E + A = C + B which may be written as eq.2a) A + E = C + B
Hint #2
In eq.4, replace A + E with C + B (from eq.2a): B + D - A = C + B - F Subtract B from both sides of the above equation: B + D - A - B = C + B - F - B which becomes eq.4a) D - A = C - F
Hint #3
In eq.3, replace C - F with D - A (from eq.4a): D + F = A + D - A which becomes D + F = D Subtract D from each side of the equation above: D + F - D = D - D which makes F = 0
Hint #4
In eq.4a, substitute 0 for F: D - A = C - 0 which becomes eq.4b) D - A = C
Hint #5
eq.5 may be written as: D - A = (B + C + D + E + F) ÷ 5 In the above equation, substitute C for D - A (from eq.4b), and 0 for F: C = (B + C + D + E + 0) ÷ 5 which becomes C = (B + C + D + E) ÷ 5 Multiply both sides of the above equation by 5: 5 × C = 5 × (B + C + D + E) ÷ 5 which becomes 5×C = B + C + D + E Subtract C from each side: 5×C - C = B + C + D + E - C which becomes eq.5a) 4×C = B + D + E
Hint #6
Substitute 0 for F in eq.4: B + D - A = A + E - 0 which becomes B + D - A = A + E Add A and E to both sides of the equation above: B + D - A + A + E = A + E + A + E which becomes eq.4c) B + D + E = 2×A + 2×E
Hint #7
Substitute 4×C for B + D + E (from eq.5a) in eq.4c: 4×C = 2×A + 2×E Divide both sides of the above equation by 2: 4×C ÷ 2 = (2×A + 2×E) ÷ 2 which becomes eq.4d) 2×C = A + E
Hint #8
Substitute 2×C for A + E (from eq.4d) into eq.2a: 2×C = C + B Subtract C from each side of the equation above: 2×C - C = C + B - C which makes C = B
Hint #9
eq.6 may be written as: A = (A + E - B + C) ÷ B Substitute 2×C for A + E (from eq.4d), and C for B in the equation above: A = (2×C - C + C) ÷ C which becomes A = (2×C) ÷ C which makes A = 2
Hint #10
Substitute 2 for A in eq.4b: D - 2 = C Add 2 to both sides of the above equation: D - 2 + 2 = C + 2 which becomes eq.4e) D = C + 2
Hint #11
Substitute 2 for A in eq.4d: 2×C = 2 + E Subtract 2 from both sides of the above equation: 2×C - 2 = 2 + E - 2 which becomes eq.4f) 2×C - 2 = E
Hint #12
Substitute 2 for A, C for B, 2 + C for D (from eq.4e), 2×C - 2 for E (from eq.4f), and 0 for F in eq.1: 2 + C + C + 2 + C + 2×C - 2 + 0 = 27 which simplifies to 2 + 5×C = 27 Subtract 2 from each side of the equation above: 2 + 5×C - 2 = 27 - 2 which makes 5×C = 25 Divide both sides by 5: 5×C ÷ 5 = 25 ÷ 5 which makes C = 5
Solution
Since C = 5, then: B = C = 5 D = C + 2 = 5 + 2 = 7 (from eq.4e) E = 2×C - 2 = 2×5 - 2 = 10 - 2 = 8 (from eq.4f) and ABCDEF = 255780