Which one is the larger : $20!$ or $2^{60}$?

Question

Which one is the larger : $20!$ or $2^{60}$ ?

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I am looking for an elegant way to solve this problem, other than my solution below. Also, solution other than using logarithm that uses the analogous inequalities below.

My solution:

Write $20!$ in prime factors and $2^{n}$:

$$ 20! = (2^{2} \cdot 5)(19)(2 \cdot 3^{2})(17)(2^{4})(3 \cdot 5)(2 \cdot 7)(13)(2^{2} \cdot 3)(11)(2 \cdot 5)(3^{2})(2^{3})(7)(2 \cdot 3) (5) (2^{2}) (3) (2) $$ $$ = 2^{18} (5)(19)(3^{2})(17)(3 \cdot 5)(7)(13)(3)(11)(5)(3^{2})(7)( 3) (5) (3) $$

so it is left to compare $2^{42}$ and $(5)(19)(3^{2})(17)(3 \cdot 5)(7)(13)(3)(11)(5)(3^{2})(7)( 3) (5) (3) $.

We write the prime factors nicely as:

$$ 3^{8}5^{4}7^{2}(11) (13)(17)(19) $$ Notice

$(3)(11) > 2^{5}$,

$(13)(5)>2^{6}$,

$(19)(7) >2^{7}$,

$17 > 2^{4}$, so we now focus on

$$3^{7}5^{3}7 = 2187(5^{3})7 > 2048(5^{3})7 = 2^{11}875 >2^{11}512 = 2^{20} $$

So we have that the prime factors is larger than $2^{42}$.

Answer 1

This answer extends the comment of zwim.

First, see Stirling approximation For factorials.

I know that as $n$ goes to $\infty$, that the geometric mean of $\{1,2,\cdots,n\}$ approaches $~\displaystyle \frac{n}{e}~$ from above. Further, as $n$ increases, the ratio between $n$ and the geometric mean of $\{1,2,\cdots,n\}$ is strictly decreasing.

I also know, from involvement with a prior (similar) problem involving $(100)!$, that even with the number as large as $(100)$, the geometric mean of $\{1,2,\cdots,100\}$ is still greater than $(40)$.

This implies that since $(20) < (100)$, the geometric mean of $\{1,2,\cdots,20\}$ must be greater than $~(20 \times 0.4).~$

This surmise makes it game over, because $2^{60} = 8^{20}$. Therefore, since the geometric mean of $\{1,2,\cdots,20\}$ is greater than $(8)$, you must have that $(20)! > 8^{20}.$

Answer 2

$$20!=2^{18}\cdot3^8\cdot 5^4\cdot 7^2\cdot 11\cdot 13\cdot 17\cdot 19$$

$$19\cdot 17\cdot 13=4199>2^{12}, \\3\cdot 11>2^5$$

So it suffices to show:

$$3^7\cdot 5^4\cdot 7^2>2^{25}.$$

Now, $5\cdot 7>2^5,$ and $3^2>2^3,$ so it suffices to showing:

$$75=3\cdot 5^2>2^{6}=64$$

This shows: $$\frac{20!}{2^{60}}=\left(1+\frac{103}{4096}\right) \left(1+\frac1{32}\right)\left(1+\frac3{32}\right)^2\left(1+\frac18\right)^3\left(1+\frac{11}{64}\right)$$

Answer 3

Here is a way to go for the solution using only elementary computations. (And building partial products that get bigger than and closer to powers of two. I wanted to use first very close approximations like $3\cdot 18\cdot 19=1026>1024$, but there is no need to be so economical at the beginning, and very generous at the end...) $$ \begin{aligned} 3\cdot 12 &= 36 \\ &> 32 =2 ^5\ ,\\ 5\cdot 13 &= 65 \\ &> 64 =2 ^6\ ,\\ 6\cdot 7\cdot 9\cdot 11 &= 6\cdot 11\cdot 7\cdot 9 =66\cdot 63=(64+2)(64-1)=64^2 +64-2 \\ &> 64^2 = (2^8)^2=2^{12}\ , \\ 15\cdot 17\cdot 14\cdot 19 &= (256-1)(16-2)(16+3)= (256-1)(256+16-6) \\ &>(256-1)(256+2)=256^2 + 256-2\\ &>256^2=(2^8)^2=2^{16}\ , \\ 10\cdot 18\cdot 20 &= 3600 \\ &>2048=2^{11}\ , \\[3mm] &\text{Putting all together:} \\[3mm] 20! &=2\cdot 4\cdot 8\cdot 16 \cdot(3\cdot 12) \cdot(5\cdot 13) \cdot(6\cdot 7\cdot 9\cdot 11) \\ &\qquad\qquad\qquad \cdot(14\cdot 15\cdot 17\cdot 19) \cdot(10\cdot 18\cdot 20) \\ &>2^{\displaystyle1+2+3+4+5+6+12+16+11} \\ &= 2^{\displaystyle60}\ . \end{aligned} $$