$1000!$ is divisible by $4^a$, Find the highest value of $a$

Question

As the question goes , I tried to solve for $2^{2a}$ first but all I got was a mess. If anyone could kindly solve this for me I would be grateful .

Edit :

Thanks for the replies. I tried the problem on my own and went with this approach,

$1000! = 1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot \dots \cdot 1000$

I want to solve for $2^{2a}$ first so I consider the even numbers only

Hence,

$$1000!=(2\cdot4\cdot6\cdot8 \dots 1000) \cdot (1 \cdot 3 \cdot 5 \cdot 7 \cdot 9 \cdot \dots 999)$$

Since, odd numbers cannot be factored by $2$, they will be ignored (in the following steps).

$$1000! = 2^{500}(1 \cdot 2 \cdot 3 \cdot \dots \cdot 500) \\ = 2^{500}(2 \cdot 4 \cdot 6 \cdot 8 \cdot 10 \cdot \dots \cdot 500) \\ = 2^{500} \cdot 2^{250} \cdot (1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot \dots \cdot 250) \\ = 2^{500}2^{250}(2 \cdot 4 \cdot 6 \cdot 8 \cdot 10 \cdot \dots \cdot 250) \\ = 2^{500}2^{250}2^{125}(1 \cdot 2 \cdot 3 \cdot 4 \cdot \dots \cdot 125)$$

I repeat this process and what i get at the end is

$2^{500} \cdot 2^{250} \cdot 2^{125} \cdot 2^{62} \cdot 2^{31} \cdot 2^{15} \cdot 2^{7} \cdot 2^{3} \cdot 2^{1} = \mathbf{2^{994}}$

Therefore, the answer to the original question would be 497

Is this process correct?

Welcome to [math.se] SE. Take a [tour]. You'll find that simple "Here's the statement of my question, solve it for me" posts will be poorly received. What is better is for you to add context (with an [edit]): What you understand about the problem, what you've tried so far, etc.; something both to show you are part of the learning experience and to help us guide you to the appropriate help. You can consult this link for further guidance. — Another User, Jan 19 '23 at 12:15
Could you determine the valuation of $1000!$ with respect to the prime $p=2$ ? Then, the rest is easy. — Peter, Jan 19 '23 at 12:22
My first thought: Same problem for $(2n)!$, small values of $n$. You get the sequence $$ 1,~3,~4,~7,~8,~10,~11,~15,~16,~18, \ldots $$ — Matti P., Jan 19 '23 at 12:25
The question itself is interesting. If the asker would show any effort at all in trying to answer it, I would probably upvote. — Adam Rubinson, Jan 19 '23 at 12:53
@AdamRubinson It would then still be an abstract duplictate. More annoying is that such questions are still answered. — Peter, Jan 19 '23 at 13:04
@everyone I am sorry that I asked the question in such a vague manner . I am new to this thing so I didnt know how to ask the questions properly . I have now edited the question and added my method of trying to solve the question. If it is correct or if its wrong , let me know kindly . — Iftekhar Alam Bhuiyan, Jan 19 '23 at 13:13
For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to. — Henrik supports the community, Jan 19 '23 at 13:25
@Peter can you provide a link to the duplicated question? If so, I'll vote to close. — Adam Rubinson, Jan 19 '23 at 13:29
Previous question about $1000!$ and base $4$: What is the maximum value of $n$ if $4^n$ divides $1000!$ without a remainder? — peterwhy, Jan 19 '23 at 13:38
@peterwhy It does. Thank you, could you check my process and see if it has any faults in it ? — Iftekhar Alam Bhuiyan, Jan 19 '23 at 13:45
No need to check my process the same process is at that question — Iftekhar Alam Bhuiyan, Jan 19 '23 at 13:49
@IftekharAlamBhuiyan I think your approach is good, until you wrote your "answer to the original question": the highest value of $a$ should be $\frac{994}2$, not the $4^{994/2}$ that you wrote. — peterwhy, Jan 19 '23 at 13:51

score 2 · Accepted Answer · answered Jan 19 '23 at 12:30

For a number $n$, denote by $v_2(n)$ the highest power of 2 which divides it, such that $v_2(1) = 0$, and $v_2(2^k) = k$ and so on.

Then it is clear that $v_2(1000!) = \sum_{n=1}^{1000}v_2(n)$.

To count the latter, there is a simple trick one can use: write $v_2$ as a sum of indicators. Write $I_k(n) = 1$ if $2^k$ divides $n$ and $0$ otherwise.

Clearly, $v_2 = \sum_{k=1}^{\infty}I_k$.

Now plug this identity into the previous sum: $$v_2(1000!) = \sum_{n=1}^{1000}v_2(n) = \sum_{k=1}^{\infty}\sum_{n=1}^{1000}I_k(n).$$

The inner summation is simply $[1000/2^k]$, hence the largest power of $2$ that divides $1000!$ is

$$[1000/2] + [1000/4] + ... + [1000/512]$$

This is something you can do by hand or plug in a calculator. The point is that it gives you a number $b$ which has the property that $2^b | 1000!$ but $2^{b+1}$ does not divide $1000!$.

The maximal $a$ for which $4^a | 1000!$ is then $[b/2]$.

Iftekhar, if you accept the answer, you should also upvote it. — user25406, Jan 19 '23 at 14:51

$1000!$ is divisible by $4^a$, Find the highest value of $a$

Edit :

1 Answers1