The least positive number such that the number of divisors of the number of divisors of the number of divisors of the number of divisors of the original number is \(3\).

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The least positive number such that the number of divisors of the number of divisors of the number of divisors of the number of divisors of the original number is \(3\).

Mathematics
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Now, I get \(72\) which is apparently wrong by doing repeated backward-working. ``` 72 => 1,2,3,4,6,8,9,12,24,36,72 | | V 12 => 1,2,3,4,6,12 | | V 6 => 1,2,3,6 | | V 4 => 1,2,4 | | V 3 ```
So is there a number smaller than \(72\) which satisfies the conditions?
60 ?

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``` 60 = 2^2 * 3 * 5 | | V 12 | | V . . . ``` OMG, so 60 is the answer?!
I get how you did the last step by doing \(12 = 2 \cdot 2 \cdot 3\ \) :-) I did the rest of the steps just like that!
Is \(60\) it?
I think it is.
60 is the least number with 12 divisors, I'll tell you how I remembered that. Gimme a min.
No, I know the divisor function. I was just doing least numbers throughout :-)
hmm.
For example, take \(2^2 3^1\). This number has \((2 + 1)(1 + 1) = 6\) divisors.
And to find the least number, you first prime factorize the number, then adjust the powers such that the least prime number gets the highest power and so on.
\[6 = 2\cdot 3 = 3\cdot 2 =6\cdot 1 =1 \cdot 6\]Now we can kinda see that it's evident how \(2^2 3^1\) is the least number. :-)
I couldn't realize that we could take a product of three primes too :-)
Do you know how divisor function works?
I had seen a very similar question on OS long time ago. That is how I could instantly say 60 ! :P Nevermind, I follow your reasoning very well. Kudos! B|
L O L

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