Puzzle for October 12, 2021 ( )
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
In eq.3, replace D with F – C (from eq.2): C + F – C = E which becomes F = E
Hint #2
In eq.4, replace F with E: B + E = A + E Subtract E from each side of the equation above: B + E – E = A + E – E which makes B = A
Hint #3
eq.6 may be written as: D = (C + E) ÷ 2 Multiply both sides of the above equation by 2: 2 × D = 2 × (C + E) ÷ 2 which becomes eq.6a) 2×D = C + E
Hint #4
In eq.6a, substitute C + D for E (from eq.3): 2×D = C + C + D which becomes 2×D = 2×C + D Subtract D from each side of the above equation: 2×D – D = 2×C + D – D which makes D = 2×C
Hint #5
Substitute 2×C for D in eq.2: F – C = 2×C Add C to both sides of the above equation: F – C + C = 2×C + C which makes F = 3×C and also makes E = F = 3×C
Hint #6
Substitute A for B, and 2×C for D in eq.5: A + A + C = 2×C – A which becomes 2×A + C = 2×C – A In the above equation, subtract C from both sides, and add A to both sides: 2×A + C – C + A = 2×C – A – C + A which makes 3×A = C Divide both sides by 3: 3×A ÷ 3 = C ÷ 3 which makes A = ⅓×C and also makes B = A = ⅓×C
Solution
Substitute ⅓×C for A and B, 2×C for D, and 3×C for E and F in eq.1: ⅓×C + ⅓×C + C + 2×C + 3×C + 3×C = 29 which simplifies to 9⅔×C = 29 Divide both sides of the above equation by 9⅔: 9⅔×C ÷ 9⅔ = 29 ÷ 9⅔ which means C = 3 making A = B = ⅓×C = ⅓ × 3 = 1 D = 2×C = 2 × 3 = 6 E = F = 3×C = 3 × 3 = 9 and ABCDEF = 113699