Convert This Number To Binary First!

Answer : There is no such $n$ .

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Proof (by contradiction) : Suppose there exists such $n$ that satisfy the constraint given, then by De Polinac's formula, $\displaystyle n! = \prod_{\text{prime }p\leq n} p^{s_p (n)}$ where $\displaystyle s_p(n) = \sum_{j=1}^\infty \left \lfloor \dfrac n{p^j} \right \rfloor$ .

In this case, we want to find the integer $n$ such that $\displaystyle n = s_2(n) = \sum_{j=1}^\infty \left \lfloor \dfrac n{2^j} \right \rfloor$ .

Let $m$ be the smallest integer such that $2^m > n$ , we can rewrite the summation as:

$n = \sum_{j=1}^\infty \left \lfloor \dfrac n{2^j} \right \rfloor = \sum_{j=1}^{m-1} \left \lfloor \dfrac n{2^j} \right \rfloor + \sum_{j=m}^\infty \left \lfloor \dfrac n{2^j} \right \rfloor$

Since $\dfrac n{2^m}, \dfrac n{2^{m+1}}, \dfrac n{2^{m+2}}, \dfrac n{2^{m+3}}, \ldots < 1$ , then the sum of floor functions of these expressions are equal to 0, thus the latter sum the in expression above is equal 0.

$n = \sum_{j=1}^\infty \left \lfloor \dfrac n{2^j} \right \rfloor = \sum_{j=1}^{m-1} \left \lfloor \dfrac n{2^j} \right \rfloor + \cancelto{0}{\sum_{j=m}^\infty \left \lfloor \dfrac n{2^j} \right \rfloor}$

Since $\left \lfloor \dfrac n{2^j} \right \rfloor \leq \dfrac n{2^j}$ , then

$\begin{aligned} n &=&\sum_{j=1}^{m-1} \left \lfloor \dfrac n{2^j} \right \rfloor \\ &=& \left \lfloor \dfrac n{2^1} \right \rfloor + \left \lfloor \dfrac n{2^2} \right \rfloor + \left \lfloor \dfrac n{2^3} \right \rfloor + \cdots \left \lfloor \dfrac n{2^{m-1}} \right \rfloor \\ &\leq & \dfrac n{2^1} + \dfrac n{2^2} + \dfrac n{2^3} + \cdots \dfrac n{2^{m-1}} \\ 1 &\leq &\dfrac1{2^1} + \dfrac1{2^2} + \dfrac1{2^3} +\cdots + \dfrac1{2^{m-1}} < \dfrac1{2^1} + \dfrac1{2^2} + \dfrac1{2^3} + \cdots + \dfrac1{2^{m-1}} + \cdots = 1 \end{aligned}$

The inequality tells us that $1<1$ , which is absurd. Hence, a contradiction.

The largest value of x, where 2^x < n, is log2(n). So x+(x-1)+(x-2)+...+1 + (n/2 - x), subtracting x from n/2 to eliminate duplicates, gives the largest exponent of 2 that divides n!. Since x(x+1)/2 is the summation of the x series above, we get (log2(n))(log2(n)+1)/2 + (n-2log2(n))/2 = ((log2(n))(log2(n)-1) + n) / 2. Therefore, (log2(n))(log2(n)-1) has to be >= n for n!/2^n to even have a possible solution. Plug that into Wolfram Alpha or solve it yourself. There is no solution. Note that I didn't use the FLOOR function above for brevity purposes.

(This isnt a solution but I didnt know where to put this) I do have a small problem with this in that 2^0|0!. I think the question should be rewritten to specify positive integers or the answer should be changed to "none of the above"

There is no solution.

Suppose that $n={ 2 }^{ a }$ .

How many twos are in the prime factorization of $(2^{ a })!$ ? We can find the maximum power of 2 that divides $(2^{ a })!$ : this power is ${ 2 }^{ 2^{ a }-1 }$ . So it's true that ${ 2 }^{ 2^{ a }-1 }$ divides $(2^{ a })!$ for every $a$ ; but ${ 2 }^{ a }\neq { 2 }^{ a }-1$

I analized only the case $n={ 2 }^{ a }$ because in this case the difference ${ 2 }^{ a }-n$ is minumum.

Let's say you have this integer $n$ . $n!$ is the product of all integers from $1$ to $n$ . First, we see that every second integer in this group is a multiple of 2. This means so far, we know that $n!$ is a multiple of $2^\frac {n} {2}$ . We also know that every second integer in this group of even integers is a multiple of 4, meaning a fourth of n integers are still even after we took out one factor of 2, so we now know that $n!$ is a multiple of $2^\frac {3n} {4}$ . If we repeat as we are doing now, we will eventually find all the factors of 2 in $n!$ because we will reach a point where, for example, we can't find the 32nd integer in the group because n is 29, meaning that we have found the highest power of 2 that $n!$ is a multiple of. You will notice that no matter how many times we repeat that process, we will never reach $n$ , only some fraction very close to $n$ such as $\frac {63n} {64}$ , meaning that no matter how large $n$ is, $n!$ can never be a multiple of $2^n$ .

I think "There is no solution" is bad wording though, as 0 is a solution, because $0! = 2^0 = 1$ .

2 raised to any power requires the solution to be not merely a multiple of 2, but all of its prime factors equal to 2.

The question is asking us what n! Has as all of its prime factors, 2.

Clearly only 2!

No math required (don't get used it on Brilliant factorial)