Question 1210676: Let f(n) denote the sum of all the positive divisors of the integer n. How many integers i exist such that 1 ≤ i ≤ 2010 and f(i) = 1 + √i + i?
Found 3 solutions by ikleyn, math_tutor2020, greenestamps:
Answer by ikleyn(53979)   (Show Source):
You can put this solution on YOUR website! .
Let f(n)  denotes the sum of all the positive divisors of the integer n.
How many integers 'i' exist such that 1 ≤ i ≤ 2010 and f(i) = 1 + √i + i?
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This formula directly indicates that 'i' is the square of a prime number.
So, the number of such 'i's is the number of primes between 1 and  = 44.83...
These prime numbers are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43. (1)
The numbers 'i' themselves are the squares  ,  ,  , . . . ,  .
Thus, the ANSWER to the problem's question is the number of primes in this list (1), i.e., 14.
Answer by math_tutor2020(3842)  (Show Source):
You can put this solution on YOUR website!
n = some integer
f(n) = the sum of the positive divisors (aka factors) of n
i = some integer between 1 and 2010. Includes both endpoints
The goal: we're looking for integers in that range where f(i) = 1 + sqrt(i) + i would be true.
Perhaps the worst way to do this would be to try all integers in the set {1,2,3,4,...,2010}
I don't recommend this method and will use something else instead.
However, for illustrative purposes, let's try a random integer such as i = 12.
12 has the factors {1,2,3,4,6,12} which sum to 28.
Then we compute the following,
f(i) = 1 + sqrt(i) + i
f(12) = 1 + sqrt(12) + 12
f(12) = 16.464 approximately
The result doesn't match with the divisor sum (28), so we can rule out i = 12 from the list.
Moreover, the result (16.464) is not an integer, so that's a faster way to rule out i = 12.
The "1+i" portion is an integer, when i happens to be an integer, so we must require sqrt(i) to be an integer as well.
This only occurs when the variable "i" is a perfect square.
Let m be a positive integer and let i = m^2, to allow this substitution:
f(i) = 1 + sqrt(i) + i
f(m^2) = 1 + sqrt(m^2) + m^2
f(m^2) = 1 + m + m^2
Whatever m is, summing the divisors of m^2 yields 1 + m + m^2
If m is prime, then the divisor list of m would be {1,m} based on what it means to be prime.
The factors of m^2 would be {1,m,m^2} when m is prime.
Summing this gives the right hand side of the equation above.
This proves that m needs to be prime.
If m was composite, then the sum would exceed 1+m+m^2, since other divisors would be added in.
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As a slight detour, let's try a prime like m = 7.
This means i = m^2 = 7^2 = 49.
f(i) = 1 + sqrt(i) + i
f(49) = 1 + sqrt(49) + 49
f(49) = 1 + 7 + 49
f(49) = 57
This is indeed the sum of the divisors of 49 aka 7^2.
So we have confirmed i = 49 works for this equation.
I'll let you try other primes. Technically you don't need to show this section to your teacher, but I think it's still good practice.
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Anyway, back to the question.
At this point you need to ask yourself: how large of a prime do I need to check?
i = 2010 is the largest you can go, which means,
i = m^2
m = sqrt(i)
m = sqrt(2010)
m = 44.833 approximately
Therefore the largest prime you need to consider is m = 43.
You can use a prime number table, or an online prime number calculator, to generate this list of primes needed
2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43
This is for the variable m.
Square each of them to get the list of "i" values that are solutions.
i = m^2 = 2^2 = 4
i = m^2 = 3^2 = 9
i = m^2 = 5^2 = 25
...etc...
up to i = m^2 = 43^2 = 1849
there are 14 numbers in that list of primes which is the number of integer solutions to f(i) = 1 + sqrt(i) + i where 1 ≤ i ≤ 2010.
Answer: 14
Answer by greenestamps(13374)  (Show Source):
You can put this solution on YOUR website!
The function f(n) is the sum of the positive integer divisors of the integer n, so f(n) is always an integer.
1 and n are integers; since f(n) is an integer and  ,  must be an integer; i.e., n must be a perfect square.
Suppose n is the square of a prime number. Then the prime factorization of n is  for some prime p. By a well-known process for finding the number of positive divisors of an integer, the number of divisors of n is  ; and those divisors are 1, n, and .
So if n is the square of a prime number, the conditions of the problem are satisfied.
So every integer i less than 2010 that is the square of a prime number is one of the values we are looking for. The square root of 2010 is between 44 and 45, so every number that is the square of a prime number less than 44 is one of the numbers we are looking for. There are 14 prime numbers less than 44, so at this point the answer to the problem is at least 14.
Now suppose n is the square of a number that is not prime. Then 1, n, and are again divisors of n. But in this case there are other divisors of n, which means  . And so numbers that are squares of numbers that are not prime can not satisfy the conditions of the problem.
So 14 is our final count of the numbers that satisfy the condition of the problem.
ANSWER: 14
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