This is too Perfect

Number Theory Level 4

Suppose for a positive number $n$ with $10$ divisors, we have

$\sum_{d|n} \dfrac{1}{d}=2$

What is the value of $n$ ?

The answer is 496.

3 solutions

Samuraiwarm Tsunayoshi
Sep 14, 2014

Multiply both sides by $n$ we get

$\displaystyle \sum\limits_{d|n} \frac{n}{d} = \sum\limits_{d|n} d = 2n$

Which is equivalent to

$\sigma (n) = 2n$

Where $\sigma (n)$ denotes divisors sum function.

From above equation, we can say that $n$ is a Perfect Number .

I used the fact that $n = q\times 2^{p-1}$ for specific prime $p, q$ such that if $n$ is increasing, $p$ will also increase.

So we know that each number has increasing number of factors. Try out some numbers and we get that $\boxed{496}$ satisfies all these conditions.~~~

A good observation about the connection to perfect numbers. Even perfect numbers are of the form $2^{p-1}(2^{p} - 1)$ for primes $2,3,5,7,13, ....$ . Since $n$ has $10$ divisors we must have $p = 5$ , and so $n = 2^{4}*31 = 496$ , which is the unique even solution as all other even perfect numbers will clearly have either less or more than $10$ divisors.

@Subrata Saha We don't know of any odd perfect numbers yet, but if one exists it has been shown that it has at least $9$ distinct prime factors, and thus well in excess of $10$ divisors. Thus $496$ is indeed the unique solution to this problem, regardless of the possible existence of any odd perfect numbers.

Brian Charlesworth - 6 years, 9 months ago

Bogdan Simeonov
Sep 15, 2014

A little hole in the problem :How do you know there doesn't exist an odd perfect number with these properties?

Check out my comment above. If there does exist an odd perfect number then it is known that it will have in excess of $10$ factors.

Brian Charlesworth - 6 years, 9 months ago

A Former Brilliant Member
Sep 14, 2014

Well,I think it is not known that 496 is the only soluiton as it is not that perfect no.s are only even

This is too Perfect

The answer is 496.

3 solutions

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