IMC 2018 Q20
20. A particular integer is the smallest multiple of 72, each of whose digits is either 0 or 1.
How many digits does this integer have?
- A 4
- B 6
- C 8
- D 10
- E 12
Intermediate Mathematical Challenge 2018, Question 20. © UK Mathematics Trust. Original paper on ukmt.org.uk
Answer: E 12
First note that 72 = 8 × 9. So all multiples of 72 are multiples of 8 and of 9. As the only digits available are 0 and 1, the required integer is a multiple of the smallest power of 10 which is a multiple of 8, namely 1000. So the last three digits of the required integer are 000. Note also that this integer must be a multiple of 9, so the sum of its digits is a multiple of 9. Therefore, the smallest multiple of 9 whose digits are 1 and 0, and which is also a multiple of 8, is 111 111 111 000. This integer has 12 digits.
Solution © UK Mathematics Trust, from the IMC 2018 solutions. Longer versions with investigations: Solutions and investigations (PDF).