Puzzle for December 24, 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.4, replace D – E with A × C (from eq.6): A × C = A – D In the above equation, substitute (A + C) for D (from eq.2): A × C = A – (A + C) which becomes A × C = –C which may be written as eq.4a) A × C = (–1) × C
Hint #2
To make eq.4a true, then either: A = –1 or C = 0 Since A must be non-negative, then: A ≠ –1 and therefore makes C = 0
Hint #3
In eq.2, replace C with 0: D = A + 0 which makes D = A
Hint #4
In eq.6, replace C with 0, and D with A: A × 0 = A – E which becomes 0 = A – E Add E to both sides of the above equations: 0 + E = A – E + E which makes E = A
Hint #5
Substitute A for D in eq.5: F – B = A ÷ A which becomes F – B = 1 Add B to both sides of the equation above: F – B + B = 1 + B which becomes eq.5a) F = 1 + B
Hint #6
Substitute (1 + B) for F (from eq.5a), and A for E in eq.3: (1 + B) = A + B + A – (1 + B) which becomes 1 + B = 2×A + B – 1 – B which becomes 1 + B = 2×A – 1 Subtract 1 from each side of the equation above: 1 + B – 1 = 2×A – 1 – 1 which becomes eq.3a) B = 2×A – 2
Hint #7
Substitute 2×A – 2 for B in eq.5a: F = 1 + 2×A – 2 which becomes eq.5b) F = 2×A – 1
Solution
Substitute 2×A – 2 for B (from eq.3a), 0 for C, A for D and E, and 2×A – 1 for F (from eq.5b) in eq.1: A + 2×A – 2 + 0 + A + A + 2×A – 1 = 32 which simplifies to 7×A – 3 = 32 Add 3 to both sides of the above equation: 7×A – 3 + 3 = 32 + 3 which makes 7×A = 35 Divide both sides by 7: 7×A ÷ 7 = 35 ÷ 7 which means A = 5 making B = 2×A – 2 = 2×5 – 2 = 10 – 2 = 8 (from eq.3a) D = E = A = 5 F = 2×A – 1 = 2×5 – 1 = 10 – 1 = 9 (from eq.5b) and ABCDEF = 580559