Determine the product of distinct positive integer divisors of .
Since we’ve traditionally required prime factorization in order to count the divisors, let’s do that here as well.
Your first instinct might have been to use a similar counting argument based on the exponents of the prime factors to say, for example, that any divisor can either have or, . Because the operative conjunction here is or, you might be tempted to say that the exponent on is just . However, this only gives the product of all the twos and the product of all the threes and so on. What it doesn’t include are things like .
Tempting as it is to approach this problem via some counting argument made on the exponents, it seems like it’s going to be a pain in the ass. Let’s have a go at it from a different direction.
Let’s look at some arbitrary divisor . By the definition of being a divisor, it must be the case that exactly one other integer is also a divisor of . In fact, the product of these two divisors will result in . Now, in the same vein of that school children story of how little gauss summed all integers less than 100 together within his head, we take the same exact reasoning in that story and apply it here.
Since , it must be the case that there are exactly divisors. With the exception of , for every divisor , another divisor also exists. Since there are such unique pairs whose product is , then their product must be . However, we left out , so we need to take it into consideration.
This gives us as the product of all divisors of .
In fact, this can be extended by defining the function , then the product of all of the digits of is just . This can be shown using the same argument as above.