Only Partially Perfect?

Computer Science Level 3

A multiplicative perfect number is a positive integer $n$ who's divisors' product totals to $n^2$ .

For example, the factors of $22$ are $1,2,11,22,~~1\cdot2\cdot11\cdot22=484$

Define the term partially-multiplicative perfect number as a positive integer who's divisor's product totals to $n^{\frac{5}{2}}$ . Find the sum of the first 4 partially-multiplicative perfect numbers.

The answer is 723.

2 solutions

Brian Charlesworth
Apr 3, 2015

I am assuming that we are looking for those positive integers $n$ whose "divisor product" equals $n^{\frac{5}{2}}.$

The first such $n$ is of course $1.$ Any other such $n$ must have $5$ divisors. Since $5$ is prime, the only way we can have $5$ divisors is if $n = p^{4}$ for some prime $p.$ The first three primes are $2,3,5,$ so the next three values of $n$ are $2^{4} = 16, 3^{4} = 81$ and $5^{4} = 625.$ Adding these to $1$ gives us a desired sum of $1 + 16 + 81 + 625 = \boxed{723}.$

@Trevor Arashiro Nice question. Regarding my initial sentence, it might be an idea to change the wording of your question to indicate the "divisor product"; as presently worded it sounds more like you are asking for the sum, which I know is not what you intended. :)

Brian Charlesworth - 6 years, 2 months ago

that is right ..... i missed that 4^4 is wrong

Louay Nasr - 6 years ago

Aryan Gaikwad
Apr 3, 2015

public static void main(String args[]){
    int sum = 0;
    for(int i = 1, n = 0; n != 4; i++)
        if(factorProduct(i) == Math.pow(i, 5.0/2)){
            sum += i; n++;
        }
    System.out.println(sum);
}

private static int factorProduct(int n){
    int prod = n;
    for(int i = 2; i <= n / 2; i++)
        if(n % i == 0) prod *= i;
    return prod;
}

Only Partially Perfect?

The answer is 723.

2 solutions

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