Puzzle for November 28, 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
Add A and D to both sides of eq.4: E – A + A + D = A + B – D + A + D which becomes E + D = 2×A + B which is the same as eq.4a) D + E = 2×A + B Add F to both sides of eq.2: D + E – F + F = A + B + C + F which becomes eq.2a) D + E = A + B + C + F
Hint #2
In eq.2a, replace D + E with 2×A + B (from eq.4a): 2×A + B = A + B + C + F Subtract A and B from both sides of the equation above: 2×A + B – A – B = A + B + C + F – A – B which simplifies to eq.2b) A = C + F
Hint #3
In eq.3, replace A with C + F (from eq.2b): F – B = C + F – C which becomes F – B = F Subtract F from both sides of the above equation: F – B – F = F – F which makes –B = 0 which means B = 0
Hint #4
In eq.1 may be written as: A + B + C + F + D + E = 28 In the equation above, replace A + B + C + F with D + E (from eq.2a): D + E + D + E = 28 which is equivalent to 2×(D + E) = 28 Divide both sides of the equation above by 2: 2×(D + E) ÷ 2 = 28 ÷ 2 which becomes eq.1a) D + E = 14
Hint #5
In eq.4a, substitute 14 for D + E (from eq.1a), and 0 for B: 14 = 2×A + 0 which means 14 = 2×A Divide each side of the above equation by 2: 14 ÷ 2 = 2×A ÷ 2 which makes 7 = A
Hint #6
Substitute 7 for A in eq.5: 7 – C = C – D Add C and D to both sides of the above equation: 7 – C + C + D = C – D + C + D which becomes 7 + D = 2×C Subtract 7 from both sides: 7 + D – 7 = 2×C – 7 which makes eq.5a) D = 2×C – 7
Hint #7
Substitute 2×C – 7 for D (from eq.5a) in eq.1a: 2×C – 7 + E = 14 In the equation above, subtract 2×C from each side, and add 7 to each side: 2×C – 7 + E – 2×C + 7 = 14 – 2×C + 7 which becomes eq.1b) E = 21 – 2×C
Hint #8
Substitute 2×C – 7 for D (from eq.5a), and 21 – 2×C for E (from eq.1b) in eq.6: C + 2×C – 7 – F = 21 – 2×C + F which becomes 3×C – 7 – F = 21 – 2×C + F Add F and 2×C to both sides of the above equation: 3×C – 7 – F + F + 2×C = 21 – 2×C + F + F + 2×C which becomes 5×C – 7 = 21 + 2×F Subtract 21 from both sides: 5×C – 7 – 21 = 21 + 2×F – 21 which becomes 5×C – 28 = 2×F Divide both sides by 2: (5×C – 28) ÷ 2 = 2×F ÷ 2 which makes eq.6a) 2½×C – 14 = F
Solution
Substitute 7 for A, 0 for B, 2×C – 7 for D (from eq.5a), 21 – 2×C for E (from eq.1b), and 2½×C – 14 for F (from eq.6a) in eq.1: 7 + 0 + C + 2×C – 7 + 21 – 2×C + 2½×C – 14 = 28 which simplifies to 7 + 3½×C = 28 Subtract 7 from both sides of the equation above: 7 + 3½×C – 7 = 28 – 7 which makes 3½×C = 21 Divide both sides by 3½: 3½×C ÷ 3½ = 21 ÷ 3½ which means C = 6 making D = 2×C – 7 = 2×6 – 7 = 12 – 7 = 5 (from eq.5a) E = 21 – 2×C = 21 – 2×6 = 21 – 12 = 9 (from eq.1b) F = 2½×C – 14 = 2½×6 – 14 = 15 – 14 = 1 (from eq.6a) and ABCDEF = 706591