We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website.
Please click on "Accept cookies" if you agree to the setting of cookies. Cookies that do not require consent remain unaffected by this, see
cookie policy and privacy policy.
DECLINE COOKIES

If we let f(n) denote the sum of all the divisors of the integer n, how many integers i exist such that 1 ≤ i ≤ 2010 and f(i)=1+ √i + i?

Guest Dec 1, 2018

#1**+2 **

Suppose we take this problem statement directly at it's word.

\(\text{The sum of the divisors of }n \text{ is }1 + \sqrt{n} + n \text{ i.e.}\\ \text{the divisors are in fact }1,~\sqrt{n},~n\)

\(\text{This implies 2 things}\\ n = p^2 \text{ for some integer }p\\ \text{all of the divisors of }n \text{ are }1, ~p,~p^2\\\)

\(\text{Thus we are looking at the squares of the prime numbers}\)

I'll let you assemble a list of the squares of the primes that are less than 2010.

You might also try to prove that the squares of the primes are the only numbers that satisfy the requirement.

Rom Dec 1, 2018

#3**+1 **

Rom is right!!

Look at the following squares of prime numbers:

2^2= 4

3^2 =9

5^2 =25

7^2 =49

11^2 =121

13^2 =169

17^2 =289

19^2 =361

23^2 =529

29^2 =841

31^2 =961

37^2 =1,369

41^2 =1,681

43^2 =1,849

As you can see, there are 14 squares of ALL the prime numbers below the sqrt(2010), or 45, that meet your 2 conditions:

1 - They are ALL between 1 and 2010.

2- ALL their divisors sum up to =1 + sqrt(i) + i

3 - As an example take any of the 14 squares above such as 29^2 =841 =1 + sqrt(841) + 841 =871(sum of its divisors)

Guest Dec 1, 2018

edited by
Guest
Dec 1, 2018