PK 2019 Q23
23. Rhona wrote down a list of nine multiples of ten: 10, 20, 30, 40, 50, 60, 70, 80, 90. She then deleted some of the nine multiples so that the product of the remaining multiples was a square number. What is the least number of multiples that she could have deleted?
- A 1
- B 2
- C 3
- D 4
- E 5
Pink Kangaroo 2019, Question 23. © UK Mathematics Trust. Original paper on ukmt.org.uk
Answer: B 2
To end up with a square product, any prime factor must occur an even number of times. Rhona cannot use 70 since the factor 7 only appears once in 70, and never in any of the other numbers.
If she uses all the other numbers she gets
\[\begin{aligned} &10 \times 20 \times 30 \times 40 \times 50 \times 60 \times 80 \times 90 \\ &= (2 \times 5) \times (2 \times 2 \times 5) \times (2 \times 3 \times 5) \times (2 \times 2 \times 2 \times 5) \times (2 \times 5 \times 5) \\ &\quad \times (2 \times 2 \times 3 \times 5) \times (2 \times 2 \times 2 \times 2 \times 5) \times (2 \times 3 \times 3 \times 5) \\ &= 2^{15} \times 3^4 \times 5^9. \end{aligned}\]
To get a square, she needs to make the powers of 2 and 5 even, say by removing 10 from the list (although removing 40 or 90 works too). Hence she needs to remove only two numbers from her list.
Solution © UK Mathematics Trust, from the PK 2019 solutions.