Puzzle for December 26, 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 E to both sides eq.3: B + E + E = A + C – E + E which becomes eq.3a) B + 2×E = A + C Add B, E, and C to both sides of eq.5: A – B + D – E + B + E + C = B – C + E + F + B + E + C which becomes A + D + C = 2×B + 2×E + F which may be written as eq.5a) A + C + D = 2×B + 2×E + F
Hint #2
In eq.5a, replace A + C with B + 2×E (from eq.3a): B + 2×E + D = 2×B + 2×E + F Subtract B and 2×E from both sides of the above equation: B + 2×E + D – B – 2×E = 2×B + 2×E + F – B – 2×E which becomes eq.5b) D = B + F
Hint #3
In eq.2, substitute B + F for D (from eq.5b): F = B + B + F which becomes F = 2×B + F Subtract F from each side of the equation above: F – F = 2×B + F – F which makes 0 = 2×B which means 0 = B
Hint #4
Substitute 0 for B in eq.2: F = 0 + D which makes F = D
Hint #5
Substitute D for F in eq.6: D – A = D ÷ D which becomes D – A = 1 Add A to both sides of the equation above: D – A + A = 1 + A which makes eq.6a) D = 1 + A which also makes eq.6b) F = D = 1 + A
Hint #6
Add E, A, and D to both sides of eq.4: D – E – A + E + A + D = A – C – D + E + A + D which becomes eq.4a) 2×D = 2×A – C + E Multiply both sides of eq.6a by 2: 2×D = 2×(1 + A) which becomes eq.6c) 2×D = 2 + 2×A
Hint #7
Substitute 2×A – C + E for 2×D (from eq.4a) in eq.6c: 2×A – C + E = 2 + 2×A In the above equation, subtract 2×A from both sides, and add C to both sides: 2×A – C + E – 2×A + C = 2 + 2×A – 2×A + C which becomes eq.6d) E = 2 + C
Hint #8
Substitute 0 for B, and (2 + C) for E (from eq.6d) in eq.3a: 0 + 2×(2 + C) = A + C which becomes 4 + 2×C = A + C Subtract C from each side of the equation above: 4 + 2×C – C = A + C – C which becomes eq.3b) 4 + C = A
Hint #9
Substitute 4 + C for A (from eq.3b) in eq.6b: F = D = 1 + 4 + C which makes eq.6e) F = D = 5 + C
Solution
Substitute 4 + C for A (from eq.3b), 0 for B, 5 + C for D and F (from eq.6e), and 2 + C for E (from eq.6d) in eq.1: 4 + C + 0 + C + 5 + C + 2 + C + 5 + C = 36 which simplifies to 5×C + 16 = 36 Subtract 16 from each side of the above equation: 5×C + 16 – 16 = 36 – 16 which makes 5×C = 20 Divide both sides by 5: 5×C ÷ 5 = 20 ÷ 5 which means C = 4 making A = 4 + C = 4 + 4 = 8 (from eq.3b) D = F = 5 + C = 5 + 4 = 9 (from eq.6e) E = 2 + C = 2 + 4 = 6 (from eq.6d) and ABCDEF = 804969