What is the largest perfect square that can be written as the product of three different one-digit positive integers?
For a positive integer to be a perfect square, its prime factors must occur in pairs.

Since only one number from the list 1, 2, 3, 4, 5, 6, 7, 8, 9 is divisible by 5 (namely 5) and only one number from the list is divisible by 7 , then neither 5 nor 7 can be one of the three integers chosen.

Therefore, we need to choose three numbers from the list 1, 2, 3, 4, 6, 8, 9 .

We note that $9 \cdot 8 \cdot 2=144=12^{2}$, which is a perfect square.

Are there perfect squares larger than 144 that can be made?

If the product of three integers is odd, then all three integers are odd. The only three integers in the list that are odd are $1,3,9$. Their product is 27 , which is not a perfect square.

Therefore, any perfect square larger than 144 that can be made must be even.

Further, note that the largest possible product of three numbers in the second list is $9 \times 8 \times 6$ or 432 , and so any perfect square that can be made is less than 432 .

The even perfect squares between 144 and 432 are $14^{2}=196,16^{2}=256,18^{2}=324$, and $20^{2}=400$.

We cannot make 196 or 400 , since no number in the list is a multiple of 7 or 5 .

We cannot obtain a product of $18^{2}=324=3^{4} 2^{2}$, since 324 has 4 factors of 3 , which would require using 3,6 and 9 , whose product is too small.

We cannot obtain a product of $16^{2}=256=2^{8}$, since if we multiply all 7 numbers from the list, there are only 7 factors of 2 in total (from $2,4,6,8$ ).

Therefore, 144 is the largest perfect square which is the product of three different one-digit positive integers.

ANSWER: 144