SOLUTION: For a positive integer n, let \tau(n) be the sum of the positive integer divisors of n. Find the number of values of n, where 1 \le n \le 25, such that \tau(n) = 1.

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Question 1210514: For a positive integer n, let \tau(n) be the sum of the positive integer divisors of n. Find the number of values of n, where 1 \le n \le 25, such that \tau(n) = 1.
Found 2 solutions by ikleyn, greenestamps:
Answer by ikleyn(53339)   (Show Source): You can put this solution on YOUR website!
.
For a positive integer n, let \tau(n) be the sum of the positive integer divisors of n.
Find the number of values of n, where 1 \le n \le 25, such that \tau(n) = 1.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~ 


Among all positive integer numbers 'n', there is only one value 'n', for which 

    tau(n) = 1.



This exclusive positive integer  'n' is  n = 1.


For all other positive integer 'n',  tau(n)  is greater than 1, which is obvious.


Therefore, there is only one positive integer 'n', for which  tau(n) = 1.


This exclusive value of 'n'  is  n = 1.

Solved, with detailed explanations.


This task isn't worth the shell of an eaten egg - so simple is it.



Answer by greenestamps(13248)   (Show Source): You can put this solution on YOUR website!


The problem as posted is either (1) trivial or (2) posted incorrectly.

The described function is the sum of the positive integer divisors of the positive integer n.

For every positive integer n greater than 1, the divisors include both 1 and n, so the sum of the integer divisors of n is at least 1+n.

So (trivially) 1 is the only positive integer n for which the sum of the divisors of n is equal to 1; therefor, the number of positive integers that satisfy the given condition is 1.

ANSWER: 1
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