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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 = 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 :-)
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|
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