Puzzle for October 16, 2019 ( )
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
Help Area
Hint #1
Add D to both sides of eq.3: F – D + D = A + E + D which becomes eq.3a) F = A + E + D Add E to both sides of eq.5: A + B + E = F – E + E which becomes eq.5a) A + B + E = F
Hint #2
In eq.5a, replace F with A + E + D (from eq.3a): A + B + E = A + E + D Subtract A and E from both sides of the above equation: A + B + E – A – E = A + E + D – A – E which simplifies to B = D
Hint #3
In eq.4, replace D with B: B + E = A + B + C Subtract B from each side of the equation above: B + E – B = A + B + C – B which becomes eq.4a) E = A + C
Hint #4
In eq.6, substitute (A + C) for E (from eq.4a): B – (A + C) = A – C which is equivalent to B – A – C = A – C Add A and C to both sides of the above equation: B – A – C + A + C = A – C + A + C which makes B = 2×A and also makes D = B = 2×A
Hint #5
Substitute 2×A for D, and (A + C) for E (from eq.4a) in eq.2: C = 2×A – (A + C) which is the same as C = 2×A – A – C which becomes C = A – C Add C to both sides of the above equation: C + C = A – C + C which means 2×C = A Divide both sides by 2: 2×C ÷ 2 = A ÷ 2 which makes C = ½×A
Hint #6
Substitute ½×A for C in eq.4a: E = A + ½×A which makes E = 1½×A
Hint #7
Substitute 2×A for D, and 1½×A for E in eq.3: F – 2×A = A + 1½×A which becomes F – 2×A = 2½×A Add 2×A to both sides of the equation above: F – 2×A + 2×A = 2½×A + 2×A which makes F = 4½×A
Solution
Substitute 2×A for B and D, ½×A for C, 1½×A for E, and 4½×A for F in eq.1: A + 2×A + ½×A + 2×A + 1½×A + 4½×A = 23 which simplifies to 11½×A = 23 Divide both sides of the equation above by 11½: 11½×A ÷ 11½ = 23 ÷ 11½ which means A = 2 making B = D = 2×A = 2 × 2 = 4 C = ½×A = ½ × 2 = 1 E = 1½×A = 1½ × 2 = 3 F = 4½×A = 4½ × 2 = 9 and ABCDEF = 241439