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This has been asked here:

https://math.stackexchange.com/questions/1801900/ionofs-problem-solving

Solving a Word Problem relating to factorisation

But they did not provide context or examples

The ionof of an integer is the integer divided by the number of factors it has. For example, $\operatorname{ionof}(18) = 3$, because $18$ has $6$ factors and $18/6 = 3$.

Show that the square of any prime is the Ionof of some integer.

Squaring prime numbers and finding Ionofs to match this, but could not find any other way to do this than trial and error.

If anyone can help, I will be greatly grateful.

Edit

Somebody said that this could be a possible duplicate of Solving a Word Problem relating to factorisation. However, that does not answer my question because the answer to that one does not talk about squares, or squares of primes at least.

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marked as duplicate by Watson, Daniel Fischer May 30 '16 at 9:49

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ The question here is indeed the same as question (d) of math.stackexchange.com/q/1760884/145141. However question (d) of that post was never answered, so I would not say that this is a duplicate. $\endgroup$ – gebruiker May 30 '16 at 9:28
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For prime $p \ne 3, \text{ ionof}(9p^2) = p^2$.

Also, $ \text{ ionof}(27\times 4) = 9$.


Why this works... The number of factors is related to the prime factorization. Each distinct prime has an exponent in that factorization, and the number of factors is the product of one more than each of those exponents. Consider $84 = 2^2.3.7 \;$. The factors of this number can be seen as the products of the three sets: $\{1,2,4\},\{1,3\},\{1,7\} = \{1,2,4,3,6,12,7,14,28,21,42,84\}$ - for a count of $12 = 3\times2\times 2$.

To get $\text{ionof}(x) = p^2$, clearly you need $p^2 \mid x$, so there will be a set of (at least) size $3$ in the process above generating the number of factors. In which case you need to divide by $3$, and so we might as well make the factor of $3$ generate another set of $3$ into the number of factors calculation by using $3^2=9$.

The factors of $9p^2$ (where $p\ne3$) are $\{1,3,9,p,3p,9p,p^2,3p^2,9p^2 \}$ for a count of $9$.

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  • $\begingroup$ Wait what? Can you go a bit more in depth pleas? $\endgroup$ – SuperNinja741 Does Gaming May 30 '16 at 9:00
  • $\begingroup$ @SuperNinja741DoesGaming - updated with more explanation... $\endgroup$ – Joffan May 30 '16 at 9:10
  • $\begingroup$ Is this the only instance where it works? $\endgroup$ – SuperNinja741 Does Gaming May 30 '16 at 9:59
  • $\begingroup$ @SuperNinja741DoesGaming This gives the construction for a resulting ionof of any prime square except $9$, and I gave a separate answer for $9$ at the top of my answer. $\endgroup$ – Joffan May 30 '16 at 10:14

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