PK 2020 Q16
16. The digits from 1 to 9 are randomly arranged to make a 9-digit number.
What is the probability that the resulting number is divisible by 18?
- A \(\dfrac{1}{3}\)
- B \(\dfrac{4}{9}\)
- C \(\dfrac{1}{2}\)
- D \(\dfrac{5}{9}\)
- E \(\dfrac{3}{4}\)
Pink Kangaroo 2020, Question 16. © UK Mathematics Trust. Original paper on ukmt.org.uk
Answer: B \(\dfrac{4}{9}\)
To be divisible by 18, the number must be divisible by 2 and by 9. The digit sum for any number formed is 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45, and hence every number is a multiple of 9. Therefore any even number formed will be divisible by 18. Of the 9 possible last digits, there are four even digits (2, 4, 6, 8), so the probability of the number being even is \(\tfrac{4}{9}\).
Solution © UK Mathematics Trust, from the PK 2020 solutions.