23. It is possible to choose, in two different ways, six different integers from 1 to 9 inclusive such that their product is a square. Let the two squares so obtained be $p^{2}$ and $q^{2}$, where $p$ and $q$ are both positive. What is the value of $p+q$ ?
A 72
B 84
C 96
D 108
E 120
23. D The product of the positive integers from 1 to 9 inclusive is $1 \times 2 \times 3 \times 4 \times 5 \times 6 \times 7 \times 8 \times 9$. As the product of prime factors, this product is $2 \times 3 \times 2^{2} \times 5 \times 2 \times 3 \times 7 \times 2^{3} \times 3^{2}=2^{7} \times 3^{4} \times 5 \times 7$. Therefore, for the product of six of the integers to be a square, it is necessary to discard 5,7 and an odd power of 2 smaller than 10 , that is 2 or 8 .
Let $p<q$. Then, discarding 8 gives $p^{2}=2^{4} \times 3^{4}$. So $p=2^{2} \times 3^{2}=4 \times 9=36$. Similarly, discarding 2 gives $q^{2}=2^{6} \times 3^{4}$. Therefore $q=2^{3} \times 3^{2}=8 \times 9=72$.
Hence $p+q=36+72=108$.