Product-perfect Numbers

Let $p(N)$ denote the product of proper divisors of $N$ , and we seek $p(N)=N$ . Consider the following cases, according to the number of distinct prime factors of $N$ :

CASE 1. $N$ has 1 prime factor. $N=p^{\alpha}$ , where $p$ is a prime number, and $\alpha$ is a positive integer. Then, $p(N) = 1\cdot p\cdot p^2\cdots p^{\alpha-1} = p^{(\alpha-1)(\alpha)/2}$ for $p(N)=N$ we have $p^{\alpha} =p^{(\alpha-1)(\alpha)/2}$ . Comparing exponents, this gives $\alpha=0,3$ , and we reject $\alpha = 0$ .This means that $\alpha=3$

CASE 2. $N$ has 2 prime factors. $N=p_1^{\alpha_1} \times p_2^{\alpha _2}$ where $p_1, p_2$ are distinct prime numbers, and $\alpha_1, \alpha_2$ are positive integers.

Case 2A. $(\alpha_1, \alpha_2) \neq (1,1)$ , then we can define $N= p_1 \times p_2 \times q$ where $q\neq 1$ . Then $P(N) \geq p_1 \times q \times p_2 \times q = p_1\times p_2 \times q^2 > N$ .

Case 2B. $(\alpha_1, \alpha_2) = (1,1)$ . Then, $p(N)= p \times q = N$ .

CASE 3. $N$ has at least 3 prime factors. $N$ can be expressed in form $p_1 \times p_2 \times q$ where $p_1 , p_2$ are primes and $q$ is any number $> 1$ . Note , in this case $p(N) \geq (p_1 \times q ) \times (p_2 \times q ) = p_1 \cdot p_2 \cdot q^2 > p_1 \cdot p_2 \cdot q =N$ . Hence, there are no solutions in this case.

Hence, the only possibilities are $N=p^3$ and $N=pq$ . So numbers will be $6,8,10,14,15,21,22,26,27,33,34,35,38,39,46$ , for a total of $\boxed{15}$ .

[Edits for clarity - Calvin]

For N to be a product perfect number it should have exactly 2 proper divisor (excluding 1). Now let those two proper divisor be 'a' and 'b'. N=a*b

Now firstly 'a' and 'b' should be prime So if a=2 then b={3,5,7,11,13,17,19,23}....so we got 8 numbers from here if a=3, then b={5,7,11,13}...we got 4 numbers from here if a=5 then b={7}..so 1 number here

So we got 13 product perfect number by now

Now if 'b' can be written as square of 'a' then also it is fine provided 'a' is prime. so if a=2 then b=4 and N=8 also if a=3 then b=9 and N=27 So total 15 numbers

6,8,10,14,15,21,22,26,27,33,34,35,38,39,46

The idea is that the product of the proper divisors grows large very fast. Let $p$ , $q$ , and $r$ be primes.

Case 1: A number has only one prime factor. Then it's $p^k$ for some integer $k$ . It is product-perfect if $p \times p^2 \times \ldots \times p^{k-1} = p^{(k-1)(k)/2} = p^k$ . The only solution is $k = 3$ , and the only numbers of the form $p^3$ less than $50$ are $8$ and $27$ .

Case 2: A number has more than one prime factor. First off, a number of the form $pq$ works, since $p \times q = pq$ . If the exponent of $p$ was any bigger, there would be too many factors of $p$ : for example, $p^2 q$ has product $p \times p^2 \times q \times qp$ .

More rigorously, if the number was $p^k q$ , the product would be $p \times p^2 \times \ldots \times p^{k} \times q \times pq \times \ldots \times p^{k-1}q$ which contains $2k$ factors of $p$ . That's too much. And it'll only get worse if you increase the power of $q$ , or add more factors like $r$ . So $pq$ is the only other possible form, and there are $13$ such numbers less than $50$ . $13 + 2 = 15$ .

We can safely exclude 1. Firstly I removed all the Primes between 1 & 50 as if p is a prime then the only divisors of p are 1 & p itself. By definition of proper divisor p fails to be product-perfect. Then I removed all the square numbers(If I consider a^2 between 1 & 50 where a is odd then clearly a^2 is not product-perfect. If a be even then a^2 has some proper divisors, the product of which exceeds a^2 & thus a^2 fails to be product-perfect). After removing the numbers our list becomes very short. After doing till this step I observed that the only product-perfect numbers among the remaining ones are those which have only two divisors other than 1 & itself & hence after finding out the product-perfect numbers among the rest I presented my result which turned out to be correct.

We can see that 6,8,10,14,15,21,22,26,27,33,34,35,38,39,46 are the required numbers. We claim that the numbers of the form pq, p and q are distinct primes, or of the form p^3, p is a prime, are the required numbers. Note that if there are four factors or more, then the product of the proper prime factors(repetition is double counted) is more than the number itself. If there are two prime distinct factors of a number z, say x, y , then xy=z, so z is a product-perfect, x not equal to y. If there are three prime factors, say a,b,c, of z, and if exactly two of them are distinct,say a,b, then the product of the proper factors is (a)(b)(ac)(bc) =z^2, which is not z. If all are distinct, then the product of the proper factors is (a)(b)(c)(ab)(ac)(bc)=z^3. thus a=b=c, and so z is the cube of a prime. also , z cannot be a prime. hence we have proved our claim.

I claim that N is product-perfect if and only if N has a prime factorization of two possible forms: N=(p^1)*(q^1), where p \neq q. That is, N is the product of exactly two primes, and the two primes are distinct.

The other form is N=(p^3). In this case, the proper divisors are 1, p, and p^2, so their product is clearly N.

Let's consider the cases where N's factorization is not of that form.

Case 1: N=(p^k), k \neq 3. If k = 1, the only proper divisor of N is 1. If k = 4, the proper divisors are 1, p, p^2, and p^3, so the product of these is p^6 > N. Clearly for k \geq 4 the product of the proper divisors will always exceed N.

Case 2: N = p q r. That is, N has 3 distinct prime factors. In this case, the proper divisors are p, q, r, pq, pr, and qr, and so their product is greater than N. Similarly if N has 4 or more prime factors.

Case 3: N = (p^2)*q. In this case the proper divisors are p, p^2, and pq, so the product of these exceeds N.

All that remains is to use the appropriate prime factorizations to list product-perfect numbers that do not exceed 50. They are:

1. 2^3 = 8
2. 2 \times 3 = 6
3. 2 \times 5 = 10
4. 2 \times 7 = 14
5. 2 \times 11 = 22
6. 2 \times 13 = 26
7. 2 \times 17 = 34
8. 2 \times 19 = 38
9. 2 \times 23 = 46
10. 3^3 = 27
11. 3 \times 5 = 15
12. 3 \times 7 = 21
13. 3 \times 11 = 33
14. 3 \times 13 = 39
15. 5 \times 7 = 35

Since the prime factorization of numbers is often key in finding numbers like these, I tried to look at the factorizations of product-perfect numbers. I found about 6 product-perfect numbers:

6 = 2 3 8 = 2 2 2 10 = 2 5 ... 26 = 2 13 27= 3 3 3 ... 33 = 3 11

By now, it's easy to see that all product-perfect number are:

a) The product of two distinct prime numbers b) The cube of a prime

Using this, I can find all product-perfect numbers below 50:

2 3, 2 5, 2 7 ... 2 23 -> 8 numbers 3 5, 3 7, 3 11, 3 13 -> 4 numbers 5*7 -> 1 number This should equal 8+3+1 = 13 numbers

Then, adding on the perfect cubes:

2^3 = 8 3^3 = 27

13+2 = 15 numbers, which would be the number of product-perfect numbers below 50.

product perfect ::it is a number which is equal to the product of all its proper divisor

we will count all the product perfect number below 50 and we will neglect all the perfect squares which come in between . 6=3 2 1 10=5 2 1 14=7 2 1 15=5 3 1 21=7 3 1 22=11 2 1 26=13 2 1 33=11 3 1 34=17 2 1 35=7 5 1 38=19 2 1 39=13 3 1 42=21 2 1 45=7 5 1 46=23 2 1 so total number of product perfect numbers below 50 is 15 so answer is 15

First find number of primes with cube less than 50 i.e. 2 (8 & 27) Number of primes greater than 2, multiply by 2 with product less than 50 i.e. 8 (3, 5, 7, 11, 13, 17, 19, 23) Number of primes greater than 3, multiply by 3 with product less than 50 i.e. 4 (5, 7, 11, 13) Number of primes greater than 5, multiply by 5 with product less than 50 i.e. 1 (7)

Total numbers = 2 +8 + 4 + 1

Note that if N is product-perfect, then either N=p^3 for a prime p or N=p*q for distinct primes p, q. There are 13 numbers which satisfy the second case (6, 10, 14, 22, 26, 34, 38, 46, 15, 21, 33, 39, 35) and 2 which satisfy the first (8, 27) for a total of 2+13=15.

Firstly, primes are not perfect, since the product of their factors is 1.

Let N be a power of a prime, say p^i. The product of its factors is then p^(i*(i-1)/2) which is equal to p^i only for i=3. Therefore, all cubes are perfect.

Now let N be a product of multiple primes: p1, p2, .., pn. There are 2^(n-1) - 1 possible proper subsets of p2, p3, .., pn. p1 multiplied with any of these is a factor of N, and therefore the power of p1 in the product is 2^(n-1) - 1. Therefore, n=2 for N to be perfect.

Finally, we see with the same method as above that N as a product of powers of multiple primes cannot be perfect.

Therefore the perfect numbers are: 2^3 3^3 2 {3, 5, 7, 11, 13, 17, 19, 23} 3 {5, 7, 11, 13} 5*{7}

Clearly, if a number is a product of 2 primes, it must be product perfect. If a number is a cube of a prime, it will also be product perfect. Hence, we only need to list the number of such numbers and count them.

Product perfect numbers mean all product divisors must be prime numbers. From this fact, 13 numbers can be found, namely 6, 10, 14, 15, 21, 22, 26, 33, 34, 35, 38, 39, 46. all of which have 3 product divisors, one of which is obviously 1 and the other 2 are prime numbers. Cubic numbers are also product perfect numbers if its cubic root is prime. That will include 8 and 27 too. e.g. 8 = 2x2x2 = 1x2x4. This gives the answer 15.

Let $A_n$ be the set of divisors of a number $n$ . By pairing up each divisor $d$ with $\frac {n}{d}$ , we see that $(\prod_{d \in A_n} d )^2 = (\prod_{d \in A_n} d ) ( \prod_{d \in A_n} \frac {n}{d} ) = \prod_{d \in A_n} { d \times \frac {n}{d} } = \prod_{d \in A_n} n = n^{\phi(n) }$ , where $\phi(n)$ represent's Euler's totient function which counts the number of divisors of $n$ . Hence, $\prod_{d \in A_n} = n^{ \frac {\phi(n)} {2} }$ .

The problem asks us to determine numbers such that $\prod_{d \in A_n} = n^2$ (recall that we must include $n$ , which is not a proper divisor). Hence, this implies that $\phi(n) = 4$ . Since $4 = 4 \times 1 = 2 \times 2$ , a number $(n)$ is product-perfect if, and only if, it has the form $n=p^3$ or $n=pq$ for distinct primes $p$ and $q$ .

For $n=p^3$ , only $8$ and $27$ are below $50$ .
For $n=pq$ , the numbers are $6, 10, 14, 22, 26, 34, 38, 46, 15, 21, 33, 39$ and $35$ .
Hence there are $15$ of them.

Suppose n is a product perfect number.Then according to definition we must have that $n$ is the product of all its divisors except $n$ itself.Multiply $n$ on both sides this means $n^2$ is the product of all divisors of $n$ .Now we know that $\prod_{d \mid n} d = n^{\dfrac{\tau(n)}{2}}= n^2$ .Compare exponents to get $\tau(n) = 4$ . Let us first note than $n$ cannot be a prime neither can it be a perfect square as this would mean $\tau(n) = 2$ and $\tau(n) = \text{odd}$ $\text{number}$ .This leaves the remaining numbers.One can brute force here onwards but however I have made it shorter.

If $\tau(n) = 4$ this means number is either of form $p_{1}^3$ or of form $p_{1}p_{2}$ .Now its simple here onwards.

The numbers as $6,8,10,14,15,21,22,26,27,33,34,35,38,39,46$ .

The actual easiest way is to see if the number has 4 factors. That is also the simplest way of stating it.

You simply have to find numbers with exactly 4 divisors.

Because if you count only proper divisors you will be left with only a pair of divisors because in multiplication process 1 is negligible.

Now to obtain numbers with 4 divisors you can use casework.

Multiply 2 with all primes except 2 to get results below 50 - $\boxed{6 , 10 , 14 , 22 , 26 , 34 , 38 , 46}$ .

Multiply 3 with all primes except 2 and 3 to get results below 50 - $\boxed{15 , 21 , 33 , 39}$ .

Multiply 5 with all larger primes to get results below 50 - $\boxed{35}$ .

Now don't forget $2^{3}$ and $3^{3}$ = $\boxed{8}$ and $\boxed{27}$ respectively.

So you get $\boxed{15}$ numbers.

In the Sieve of Eratosthenes,

1) One can eliminate all the primes as the primes have only one as their divisors. Example : 17 : 17,1

2) One can eliminate all the multiples of 4, 6, 8, 10 as those numbers also have 2 as their divisor.Example : 28 : 1,4,7 also 2 12 : 1,3,4 also 2

3) One can cancel all the squares too.

The remaining numbers are product-perfect numbers.

We simply exhaust all numbers with 2 distinct proper prime divisors. Find the smallest and largest prime such that their product is less than or equal to 50. These two numbers are 2 and 23. Listing all prime numbers in between the two we get:

2, 3, 5, 7, 11, 13, 17, 19, 23

We then take the product of the numbers two at a time such that the product is less than 50. We would obtain the following values:

6, 10, 14, 15, 21, 22, 26, 33, 34, 38, 39, 46

Notice that prime perfect cubes (cubes of prime numbers) are also product perfect numbers because the given a prime perfect cube p^3, its proper divisors are 1, p, and p^2. Thus, in addition to our first set of numbers, we also have:

8, 27

Let us not forget the trivial case 1. This gives us a total of 15.