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[tubc]

Tìm điều kiện để $n!$ chia hết cho $n^3$

Cám ơn các bạn nhìu.

[nguyenta98]

Tìm điều kiện để $n!$ chia hết cho $n^3$

Cám ơn các bạn nhìu.

Lời giải của mình trên Mathlink

We have $n^2|(n-1)!$
If $n$ is a prime then we can easily get contradiction
If $n=2p$ with $p$ is a prime and $p\geq 3$ then $4p^2|(n-1)!$
We have $4|(n-1)!$ but $n=2p$ with $p$ is a prime so all numbers which are less than or equal to $n-1$ is $p$
Thus $p||(n-1)!$ so False!
If $n\neq 2p$ and $n$ isn't a prime then let $n$ be $p^k.l$ with $gcd(l,p)=1$
Case 1: $n$ is odd then $p\geq 3$
First if $l=1$ then $n=p^k$ thus $p^{k-1}<n-1<p^k$
But if $n=9=3^2$ then gives contradiction so when $p=3$ then $k\geq 3$
Then $v_p((n-1)!)=[\dfrac{n-1}{p}]+...+[\dfrac{n-1}{p^{k-1}}]$
$[\dfrac{p^k-1}{p}]+...+\dfrac{p^k-1}{p^{k-1}}]$
$[\dfrac{p^k-p+p-1}{p}]+...+[\dfrac{p^k-p^{k-1}+p^{k-1}-1}{p^{k-1}}]$
$p^{k-1}+p^{k-2}+...+p-(k-1)$
Then we need to prove $p^{k-1}+p^{k-2}+...+p-(k-1)\geq 2k$
$\Rightarrow p^{k-1}+...+p+1\geq 3k$
$\Rightarrow p^k-1\geq 3k(p-1)$
Notice that $n$ isn't a prime then $k\geq 2$ and we have proven that when $p=3$ then $k\geq 3$
If $k=2$ thus $p\neq 3 \Rightarrow p\geq 5$ then $p^2-1\geq 6(p-1)$ (this is true for $p\geq 5$)
So we can prove by induction that $p^k-1\geq 3k(p-1)$ so done
If $l\geq 2$ this is similar to $l=1$ (much more easier)
Thus if $n=p_1^{a_1}...p_t^{a_t}$ then $p_i^{2a_i}|(n-1)!$ and $gcd(p_i^{2a_i},p_j^{2a_j})=1$ then $n^2|(n-1)!$
Case 2: $n$ is even
If $n=2^k$ then $2^{k-1}<n-1<2^k$
Then $v_2((n-1)!)=[\dfrac{n-1}{2}]+...+[\dfrac{n-1}{2^{k-1}}]$
$[\dfrac{n-1}{2^i}]=[\dfrac{2^k-1}{2^i}]=[\dfrac{2^k-2^i+(2^i-1)}{2^i}]=2^{k-i}-1$
Thus $[\dfrac{n-1}{2}]+...+[\dfrac{n-1}{2^{k-1}}]=2^{k-1}-1+2^{k-2}-1+...+2{k-(k-1)}-1=(2^{k-1}+...+2)-(k-1)=2^k-4-(k-1)$
Thus we have to prove $v_2((n-1)!)\geq 2k \Rightarrow 2^k-4-(k-1)\geq 2k \Rightarrow 2^k\geq 3k+3$ and this is true for only $k\geq 4$
If $n=2^k.l$ with $l>1$ and $l$ is odd thus $l\geq 3$
If $k=1$ then $n=2l$ but $n$ doesn't have form of $2p$ with $p$ is a prime so $l$ is an odd composite prime then $l\geq 9$ so $n\geq 9$, we have $n=2l$ and from above, $l=p_1^{2a_1}...p_t^{2a_t}$ with $p_i$ is odd prime then $l|(n-1)!$ and $(n-1)!$ is divisible by $4$ for $n\geq 4$ then $4l(n-1)!$ thus $4l^2|(n-1)!$ then $n^2|(n-1)!$
If $n\geq 2$ then $2^k<n-1$ $v_2((n-1)!)=[\dfrac{n-1}{2}]+...+[\dfrac{n-1}{2^{k}}]$ and similar, it's equal to $(2^k-2).l-k$
So we have to prove $(2^k-2).l-k\geq 2k \Rightarrow (2^k-2).l\geq 3k$ with $l\geq 3$ and $k\geq 2$ and this is obviously true (induction!)

Answer
$i)-- n=\prod(p_i^{a_i})$ with $p_i$ is an odd prime and $n\neq p$ which $p$ is a prime and $n\neq 9$
$ii)-- n=2^k$ with $k\geq 4$
$iii)-- n=2^k.l$ with $l$ is a odd composite and $k\geq 1$
$iv)-- n=2^k.l$ with $l$ is a prime and $k\geq 2$

P/S I think Rust has made a mistake, you can check my answer, $n=2p$ with $p$ is a prime cannot be true !