What is the possible value for $n$, where $2^n$ is the most near value to $50!$

Question

What is the possible value for $n$, where $2^n$ is the most near value to $50!$.

(I've tried simplifying numbers like $2 = 2^1, 4 = 2^2$, etc., but the value for $n$ which I obtained is nowhere close to the $50!$)

Answer 1

Compute $$\log_2(50!)$$ Using properties of logarithm we can compute it as $$\sum_{n=1}^{50}\log_2(n)$$ After wolframAlpha we can deduce, that this sum is $$a\approx 214.2$$ Now we have two possible values of $n$: $\lfloor a\rfloor =214$ and $\lceil a\rceil =215$. Let's compute differences between $2^n$ and $50!$: $$2^{214}-50! \approx -4\cdot 10^{63}\\2^{215}-50! \approx 2\cdot 10^{64}$$ As we can see, $50!$ is between these two values and closer to $2^{214}$, thus $$n=214$$

Answer 2

If you want to calculate without using Wolfram alpha

We can write, $$2^n=50!$$ $$n\ln 2=(\ln 50!)$$ $$n\ln 2 =(\sum_{k=1}^{50} \ln k)$$ $$n\ln 2=(50\sum_{k=1}^{50}\frac{( \ln k-\ln 50)}{50}+\sum_{k=1}^{50} \ln 50)$$ $$n\ln 2=(50\sum_{k=1}^{50}\frac{( \ln k/50)}{50}+\sum_{k=1}^{50} \ln 50)$$ Using the idea of reimann integrals, $$n\ln 2=(50\int_{1/50}^{1}\ln x dx+50\ln 50)$$

$$n\ln 2=(50(-1-\frac{\ln 1/50}{50}+\frac{1}{50}+50\ln 50)$$ $$n\ln 2=(-50+\ln {50}+1+50 \ln 50)$$ $$n\ln 2=(150.5)$$ $$n=\frac{150.51}{\ln 2}$$ $$n=217.14\approx 217$$

This is much closer to the actual answer =214

Answer 3

The conceptually easiest method is to compute $50!$ explicitly, then convert it in base 2 by successive divisions. The result has 215 binary digits

10010011110111010111100100101100001111011010010011110011011000000101011000111101111010011110010100011010001100110101000010011110101100101110011101000011101001011000111100000000000000000000000000000000000000000000000

One sees that $n = 214$ because $|50! - 2^{214}|< 2^{212}$

Answer 4

You can get a pretty good estimate without tables or assistance as follows:

Up to $50$ there are $25$ even numbers, $12$ divisible by $4$, $6$ divisible by $8$, $3$ divisible by $16$ and $1$ divisible by $32$. We have $25+12+6+3+1=47$.

There are $16$ multiples of $3$, $5$ multiples of $9$ and $1$ multiple of $27$. Then $16+5+1=22$

There are $10$ multiples of $5$ and $2$ of $25$ - total $12$

There are $7$ multiples of $7$ and one of $49$ - total $8$

There are $4$ multiples of $11$, $3$ of $13$, $2$ of $17, 19, 23$ and the other primes less than $50$ appear just once.

$50!$ =$2^{47}\cdot 3^{22}\cdot 5^{12}\cdot 7^8\cdot 11^4\cdot 13^3\cdot 17^2\cdot 19^2\cdot23^2\cdot 29\cdot 31\cdot 37\cdot 41 \cdot 43\cdot 47$

So we have power $47$ from the first factor. $3^5=243$ is about $2^8=256$, (five percent error) so $3^{20}$ is about twenty percent less than $2^{32}$, and we have a factor of $9$ still to count. We can take this as $2^3$ with an error of twelve percent.

$5^3=125$ is pretty much $2^7=128$, so for $5^{12}$ we can estimate $2^{28}$ percentage error about $2.5\times 4= 10$.

$7\times 11\times 13 =1001$ and this is about $2^{10}=1024$ so we take this three times and account for a factor of $2^{30}$ less about six percent. This leaves $7^5\times 11$.

$17^2$ is about $16^2=2^8$ plus about twelve percent.

$7\times 19=133 \approx 128=2^7$ so we can take this twice as $2^{14}$, with error about four percent. We have $7^3$ left to account for, but $7^2\times 11=539\approx 512=2^9$ with an error about five percent.

Now we have $7\times 37=259\approx 2^8$ with an error about one percent.

Take stock - we have extracted total power $47+32+3+28+30+8+14+9+8=179$, the percentage error is estimated as $20-12+10-6-12-4+5-1=-2$. So we have $$50!\approx 2^{179}\cdot 23^2\cdot 29\cdot 31\cdot 41 \cdot 43\cdot 47\cdot 1.02$$where the $1.02$ comes from the percentage error.

Of course $31\approx 32=2^5$, error three percent. Bracketing $32$ we look at $23\times 43=989\approx 2^{10}$ error three and a half percent, and $23\times 47=1081\approx 2^{10}$ error five and a half percent, so these give exponent $25$ and underestimate by about one percent.

Finally $29\times 41=1189 \approx 2^{10}$ with the rather larger error of about eighteen percent.

So finally we have $50!\approx 2^{214}\times 1.21$

Answer 5

Nobody really mentioned the Stirling's formula, so that's what I'm gonna do. I'll show how it can be used for getting a quick and dirty approximation of the result. For large $n$, it says that $$ n! \approx \sqrt{2\pi n} \left(\frac n e\right)^n. $$ As some of the other answers clarify, you need to take the binary logarithm of this. Let's do that: $$\log_2 \left[ \sqrt{2\pi n} \left(\frac n e\right)^n \right] = \frac 1 2 (1 + \log_2 \pi + \log_2 n) + n(\log_2 n - \log_2 e).$$ Now we need to approximate those logarithms. $\pi$ and $e$ are both somewhere between $2$ and $4$, so we just pretend that $\log_2 \pi \approx \log_2 e \approx \frac 3 2$. Similarly, n is roughly "halfway" between 32 and 64, so let's say that $\log_2 50 = 5 + \frac 1 2$. Plugging in, we get $$\log_2 50! \approx \frac 1 2 \left[1 + \left(1 + \frac 1 2\right) + \left(5 + \frac 1 2\right)\right] + 50 \cdot \left[\left(5 + \frac 1 2\right) - \left(1 + \frac 1 2\right)\right] =\\ = \frac 1 2 \cdot 8 + 50 \cdot 4 = 204.$$

I did this just with pencil and paper. No calculators, computers, tables and so on. (That's why I approximated the logarithms as something + ½.) I think that for a 30 sec approximate calculation, we got a nice result (it's only 5% off).

Edit: I just checked with the calculator. In fact, $\log_2 50 \approx 5.64$ and $\log_2 e \approx 1.44$, so $(\log_2 50 - \log_2 e)$ is in fact 4.2. This $0.2 \cdot 50$ exactly makes the remaining 10.