What is the maximum value of $n$ if $4^n$ divides $1000!$ without a remainder?

Question

If $1000!$ is divided by $4^n$ with a remainder 0, what is the highest possible value of $n$?

I placed 2, 3, 4, etc value in $n$ but didn't found any possible $4^n$. Moreover I have seen that only $4^1$ can divide 1000! without remainder. Is there any way to prove this or there are some possible $n$?

@giannispapav I understood it. But can you please tell how can I get the highest value of $n$? I don't think calculator can find it. — Shromi, Mar 03 '19 at 13:12

rtybase · Accepted Answer · 2019-03-03T13:22:07.560

4

According to Legendre's Theorem (which is a very useful tool for such problems) $$\nu_p(n!)=\frac{n-s_p(n)}{p-1}$$ With $p=2$ and $n=1000$ $$\nu_2(1000!)=1000-s_2(1000)$$ Also $(1000)_{10}=(1111101000)_2$ so $s_2(1000)=1+1+1+1+1+0+1+0+0+0=6$, thus $$\nu_2(1000!)=1000-6=994$$ or $$2^{994}=4^{497} \mid 1000!$$

edited Mar 03 '19 at 13:22

answered Mar 03 '19 at 13:15

rtybase

16,907

1123581321 · Answer 2 · 2019-03-03T18:08:32.317

3

For a solution by hand: $1000!$ has

$500$ multiples of $2^{1}$

$250$ multiples of $2^{2}$

$125$ multiples of $2^{3}$

$62$ multiples of $2^{4}$

$31$ multiples of $2^{5}$

$15$ multiples of $2^{6}$

$7$ multiples of $2^{7}$

$3$ multiples of $2^{8}$

$1$ multiples of $2^{9}$

So $2^{1+3+7+15+...+500}=2^{994}=4^{497}|1000!$

edited Mar 03 '19 at 18:08

answered Mar 03 '19 at 13:17

1123581321

5,108

The only thing missing is $2^0=4^0=1$ but I guess it doesn't matter, being an identity. – poetasis Mar 03 '19 at 13:21
@poetasis But $2^0$ doesn't contribute a nontrivial power of $2$, does it? Could you explain it a little bit more? – 1123581321 Mar 03 '19 at 13:24
1

Shouldn't it be $125$ multiples of $2^3$? – Barry Cipra Mar 03 '19 at 13:48
1

@ giannispapav $2^0$ contributes a trivial power of two so it really doesn't matter. – poetasis Mar 04 '19 at 00:38
@giannispavav can you please explain me how the exponent of 2 is related to n? Also How there are 500 multiples of $2^1$ in 1000! – Shromi Jun 02 '19 at 12:20
1

@Shromi $1000!=1\cdot2\cdot3\cdot4\cdot...\cdot999\cdot1000=(1\cdot3\cdot...\cdot999)\cdot(2\cdot4\cdot...\cdot1000)$ – 1123581321 Jun 03 '19 at 11:41
1

@giannispavav thanks for explaining this easily. – Shromi Jun 03 '19 at 12:09

What is the maximum value of $n$ if $4^n$ divides $1000!$ without a remainder?

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