How Many Quintessential Numbers Satisfy This Property?

Number Theory Level 1

The sum of all the positive divisors of this number is twice the original number.

Which of the following number satisfy the property above?

4 5 6 7

3 solutions

Swapnil Das
Aug 25, 2016

Relevant wiki: Sum of Factors

The positive divisors of $6$ are $1$ , $2$ , $3$ and $6$ . Their sum is $1+2+3+6=12= 6 \times 2$ .
Such numbers are called perfect numbers .

$\text{The title of the problem is fairly interesting, being one the famous unsolved problems of mathematics, till date.}$

Moderator note:

Yes, for completeness, one should show that the other 3 numbers listed does not satisfy this criteria.

You might also notice the following observations:

If $p$ is a prime number , then the sum of all the positive divisors of $p$ is $p+1$ .
The sum of all the positive divisors of a power of 2, $2^n$ is equal to $2^{n+1} - 1$ .

You got the reference! Well doneeeeeeeee

Pi Han Goh - 4 years, 9 months ago

Thanks a lot :)

Swapnil Das - 4 years, 9 months ago

why 4 and 5 cannot include their..?

A Former Brilliant Member - 4 years, 9 months ago

cannot include their what?

Pi Han Goh - 4 years, 9 months ago

I got the question correct but when you say 'odd numbers' in the title, I was thinking that no known odd numbers satisfy being a perfect number. Perhaps an edit on the title to stop any confusion?

Sharky Kesa - 4 years, 9 months ago

Sure, go ahead and change the title

Pi Han Goh - 4 years, 9 months ago

@Pi Han Goh – What do you think of Quintessential?

Sharky Kesa - 4 years, 9 months ago

@Sharky Kesa – Sure. It works!

Pi Han Goh - 4 years, 9 months ago

Brock Brown
Sep 3, 2016

Python 2.7:

# cracking a nut with a sledgehammer
def property(n):
    total = n
    for i in xrange(1,n/2+1):
        if n%i == 0: total += i
    return total == 2*n
for n in (4, 5, 6, 7):
    if property(n):
        print "Answer:", n

This is the perfect solution.

Pi Han Goh - 4 years, 9 months ago

Anandmay Patel
Aug 26, 2016

Just had an idea:The first 2 options are eliminated without any calculation as they are prime.If we consider a function f,such that it relates all natural numbers to their respective 'sum of all divisors',then f(p)=p+1,for every prime p.So, as primes can not satisfy this property,many odds are deleted to examine them for the property:)