Puzzle for November 1, 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 positive integer.
Scratchpad
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Hint #1
In eq.3, replace D with A + B (from eq.1): B + A + B = A + F which becomes A + 2×B = A + F Subtract A from both sides of the above equation: A + 2×B – A = A + F – A which makes eq.3a) 2×B = F
Hint #2
In eq.4, replace A + B with D (from eq.1): D + E – A = D + C In the above equation, subtract D from each side, and add A to each side: D + E – A – D + A = D + C – D + A which becomes eq.4a) E = C + A
Hint #3
eq.6 may be written as: A + (B × C) = D + F + E In the equation above, substitute C for D + F (from eq.2), and C + A for E (from eq.4a): A + (B × C) = C + C + A which becomes A + (B × C) = 2×C + A Subtract A from both sides: A + (B × C) – A = 2×C + A – A which becomes B × C = 2×C Divide both sides by C: B × C ÷ C = 2×C ÷ C which makes B = 2
Hint #4
Substitute 2 for B in eq.3a: 2×2 = F which means 4 = F
Hint #5
Substitute 2×B for F in eq.2: eq.2a) C = D + 2×B In eq.2a, replace D with A + B (from eq.1): C = A + B + 2×B which becomes eq.2b) C = A + 3×B
Hint #6
Substitute (A + B) for D (from eq.1), and A + 3×B for C (from eq.2b) in eq.5: A + (B × (A + B)) = A + 3×B which may be written as A + B×A + B×B = A + 3×B Subtract A from both sides of the above equation: A + B×A + B×B – A = A + 3×B – A which becomes eq.5a) B×A + B×B = 3×B
Hint #7
Substitute 2 for B in eq.5a: 2×A + 2×2 = 3×2 which becomes 2×A + 4 = 6 Subtract 4 from each side of the above equation: 2×A + 4 – 4 = 6 – 4 which makes 2×A = 2 Divide both sides by 2: 2×A ÷ 2 = 2 ÷ 2 which makes A = 1
Hint #8
Substitute 1 for A, and 2 for B in eq.2b: C = 1 + 3×2 which becomes C = 1 + 6 which makes C = 7
Hint #9
Substitute 1 for A, and 2 for B in eq.1: D = 1 + 2 which makes D = 3
Solution
Substitute 7 for C, and 1 for A in eq.4a: E = 7 + 1 which makes E = 8 and makes ABCDEF = 127384