Why is 4096 the smallest number with 13 factors. What does 2^n have to do with it? I understand in order to get an odd number of factors you have to use a perfect square.
Thanks for anything you can add to the picture.
this seems like a follow up to this thread. you should have asked this question there
powers of 2 would yield the smallest number. since powers of 1 would only yield 1 and thus would not reach 4096, and powers of 3 (or more, or a product with numbers greater than 2) would be greater than corresponding powers of 2 and so would not be minimum
Hello, reagan3nc!
There is a theorem that might help explain it.Why is 4096 the smallest number with 13 factors?
What does 2^n have to do with it?
I understand in order to get an odd number of factors you have to use a perfect square.
Given: .$\displaystyle N \:=\:p^a\cdot q^b\cdot r^c\,\cdots$ . (prime factorization)
. . the number of factors of $\displaystyle N$ is: .$\displaystyle d(N) \:=\:(a+1)(b+1)(c+1)\cdots$
. . . . Add one to each exponent and multiply.
Example: .$\displaystyle N \:=\:6125 \:=\:5^3\cdot7^2$
. . . $\displaystyle d(6125) \:=\:(3+1)(2+1) \:=\:12$ factors.
We want a number $\displaystyle K$ so that: .$\displaystyle d(K) \:=\: 13 \:=\:12 + 1$
Hence, $\displaystyle K$ has a prime factorization with an exponent of 12: .$\displaystyle K \:=\:p^{12}$
. . And the smallest $\displaystyle K$ occurs when $\displaystyle p = 2.$
Got it?