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
23. B 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
$$
\begin{aligned}
10 & \times 20 \times 30 \times 40 \times 50 \times 60 \times 80 \times 90 \\
= & (2 \times 5) \times(2 \times 2 \times 5) \times(2 \times 3 \times 5) \times(2 \times 2 \times 2 \times 5) \times(2 \times 5 \times 5) \\
& \times(2 \times 2 \times 3 \times 5) \times(2 \times 2 \times 2 \times 2 \times 5) \times(2 \times 3 \times 3 \times 5) \\
= & 2^{15} \times 3^{4} \times 5^{9} .
\end{aligned}
$$

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.