Puzzle for December 1, 2021 ( )
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.5, substitute (C + D) for E (from eq.3), and (A + C) for F (from eq.2): C – D = A – (C + D) – (A + C) which becomes C – D = A – C – D – A – C which becomes C – D = –2×C – D Add D and 2×C to both sides of the above equation: C – D + D + 2×C = –2×C – D + D + 2×C which makes 3×C = 0 which means C = 0
Hint #2
In eq.2, replace C with 0: F = A + 0 which makes F = A
Hint #3
In eq.3, replace C with 0: E = 0 + D which makes E = D
Hint #4
eq.6 may be written as: B = (A + C + D + E + F) ÷ 5 Multiply both sides of the above equation by 5: 5 × B = 5 × (A + C + D + E + F) ÷ 5 which becomes eq.6a) 5×B = A + C + D + E + F
Hint #5
eq.1 may be written as: A + C + D + E + F + B = 24 Substitute 5×B for A + C + D + E + F (from eq.6a) in the above equation: 5×B + B = 24 which makes 6×B = 24 Divide both sides by 6: 6×B ÷ 6 = 24 ÷ 6 which makes B = 4
Hint #6
Substitute 4 for B in eq.4: D – 4 = A + 4 Add 4 to both sides of the above equation: D – 4 + 4 = A + 4 + 4 which makes D = A + 8 and also makes eq.4a) E = D = A + 8
Hint #7
In eq.1, substitute 4 for B, 0 for C, A + 8 for D and E (from eq.4a), and A for F: A + 4 + 0 + A + 8 + A + 8 + A = 24 which simplifies to 4×A + 20 = 24 Subtract 20 from each side of the above equation: 4×A + 20 – 20 = 24 – 20 which makes 4×A = 4 Divide both sides by 4: 4×A ÷ 4 = 4 ÷ 4 which makes A = 1 and also makes F = A = 1
Solution
Substitute 1 for A in eq.4a: E = D = 1 + 8 which makes E = D = 9 and makes ABCDEF = 140991