# Which is bigger: a googolplex or $10^{100!}$

A googol is defined as $$10^{100}$$ Let x = $$10^{100}$$ A googolplex is defined as $$10^{x}$$

Which is bigger: a googolplex or $$10^{100!}$$

I only know that: $$100! = 1×2×3×...×98×99×100$$ $$10^{100} = 10×10×10×...×10×10×10$$

I think its easier to approach if I only compare the exponents, because they both have the same base $$10$$ anyways, but I don't know how to show which is bigger from $$100!$$ and $$10^{100}$$

• Do you know how to find the exponent of a prime $p$ in $n!$ ? It is known as Legendre's Thm. Exponent of $p$ in $n!$ is given by $ν_p (n!)=\sum_{k=1}^{\infty}\lfloor\frac{n}{p^k}\rfloor$
– Sam
Commented Oct 29, 2019 at 17:13

Both $$10^{100}$$ and $$100!$$ have 100 terms. Observe that

$$(101-k)×k=101k-k^2\geq 100=10^2$$ for $$k=1,...,50$$ and equality holds only for $$k=1$$.

Thus you can put togheter 2 by 2 the terms in $$100!$$ so that their product la bigger than $$10^2$$. Thus $$100!>10^{100}$$.

Well, if $$100! > 10^{100}$$ then $$10^{100!}$$ is bigger. If $$10^{100} > 100!$$ then a googolplex is bigger.

SO which is bigger $$100!$$ or a googol?

$$100! = (1*2*....*10)*(11*...*20)*(21*....*30)*......*(91*....100)$$

$$> (1*1*....*1)*(10*10*....*10)*(20*20*...*20)*......*(90*90*...*90)$$

$$= 1^{10}\times 10^{10}\times 20^{10}\times.... \times 90^{10}$$

$$=(10^{10})\times (2^{10}*10^{10})\times.... \times (9^{10}*10^{10})$$

$$= (10^{10}*10^{10}*10^{10}*....*10^{10})\times (2^{10}*3^{10}*.....*9^{10})$$

$$=(10^{90})\times (2^{10}*3^{10}*4^{10}*5^{10}......*9^{10})$$

$$> (10^{90})\times (2^{10}*5^{10})$$

$$= 10^{90}\times 10^{10}$$

$$= 10^{100}$$.

So $$100! > 10^{100}$$ and

$$10^{100!} > 10^{10^{100}}$$

=====

Nother way of thinking of it:

$$100!$$ has $$100$$ "pieces" from $$1$$ to $$100$$. And $$10^{100}$$ has $$100$$ pieces all equal to $$10$$. The $$9$$ pieces $$1$$ to $$9$$ are all less than $$10$$ and $$90$$ of the pieces $$11... 100$$ are all larger than $$10$$.

So the question is: Do the $$90$$ pieces larger than $$10$$ "overwhelm" the product so $$100! > 10^{100}$$; or do the $$9$$ pieces of googol that are $$10$$ overwhelm the pieces of $$100!$$ that are less than $$10$$.

Another way of puttng this is:

$$(1*2*....*9)*10*(11*.....*100) <,=,> (10*10*...*10)*10*(10*....*10)\iff$$

$$\frac {11*.....*100}{10*10*....*10} <,=,> \frac{10*10*...*10}{1*2*3*....*9}$$.

Now my intuition says the $$1,2,.....9$$ are so insignificant and few compared to the many $$11,...., 100$$.

On the google side we have everything having a geometric average of $$10$$.

To "make up" for how small the $$1$$ is, we can pair it with the $$100$$ so to get $$1*100 = 10*10$$. And now the $$1$$ has been "smoothed out".

We can pair the $$2$$ with the $$50$$ to get $$2*50 =10*10$$ and that has been evened out.

We can't pair the $$3$$ with $$33\frac 13$$ but if we pair it with $$34$$ we get $$3*34 > 10*10$$ so weighting is to $$100!$$ advantage.

And so on... pair $$4*25=10*10$$ andd $$5*20 =10*10$$ and $$6*17 > 10*10$$ and $$7*15 > 10*10$$ and $$8*13 > 10*10$$ and $$9*12 > 10*10$$.

.... to put this together....

$$(1*100)*(2*50)*(3*34)*(4*25)....(9*12) > 100^9 = 10^{18}$$.

$$10*11*14*16*18*19 > 10^6$$

$$21*22*23*24*26*27*28*29*30 >10^9$$

$$31*32*33*35*....*39*40> 10^9$$

$$41*....*49>10^9$$

$$51*.....*99>10^{49}$$ and so $$1*2*....*100 > 10^{18+6+9+9+9+49}=10^{100}$$.

The question boils down to whether $$10^{100}$$ is greater than $$100!$$ (same bases). But by Stirling's approximation, $$100!\approx\sqrt{200\pi}(100/e)^{100}$$ and $$100/e>33>10$$ and $$\sqrt{200\pi}<30$$ so $$100!>10^{100}$$. Hence $$10^{100!}$$ is the bigger number.

• I need to show how you got that approximation. I don't understand how you got that number. Commented Oct 29, 2019 at 17:08
• @ViciCIT Done.${}$ Commented Oct 29, 2019 at 17:14

You can even check this with google:

$$100!>10^{100}$$

By regrouping terms in $$100!$$, using some crude inequalities, and carefully counting the number of terms involved, we have

\begin{align} 100! &=100\cdot99\cdot98\cdots11\cdot10\cdot9\cdots2\cdot1\\ &=(100\cdot1)(99\cdot2)\cdots(92\cdot9)\times(91\cdot90\cdots11\cdot10)\\ &\gt(100)(100)\cdots(100)\times(10\cdot10\cdots10\cdot10)\\ &=100^9\cdot10^{82}\\ &=10^{100}\\ &=\text{googol} \end{align}

Therefore $$10^{100!}\gt10^\text{googol}=\text{googolplex}$$

(Remark: The "$$\times$$" symbol's role here is purely visual, to put a little extra separation between things that are treated differently. The answer, in general, is very similar to Alberto Saracco's.)

• Can you explain what you did in the second and third line? I understand everything else Commented Oct 29, 2019 at 19:27
• @ViciCIT, in the second line I moved the numbers $9$ to $1$ from the right hand end of the product and paired them up with $92$ to $100$; in the third line I used the fact that the paired products are all at least $100$ while the rest of the terms are all at least $10$ (and most of them are quite a bit larger). The trickiest part of the whole thing is to count that there are $82$ numbers that get replaced with $10$'s. (It's fairly easy to see you there are $9$ numbers that get replaced with $100$'s, because you can see the numbers $1,2,\ldots,9$.) Commented Oct 29, 2019 at 20:15

Consider that $$\log_{10}(100!) = \sum_{i=1}^{100}\log_{10}(i) \ge \sum_{i=32}^{100}\log_{10}(i)\ge \sum_{i=32}^{100}\frac{3}{2} = 103.5 > 100$$ Now exponentiate both sides twice to get $$10^{100!} > 10^{10^{100}}$$ (Note that if you want a better estimate for the size of $$10^{100!}$$, the logarithmic sum can be computed exactly fairly easily, giving $$\log_{10}(100!) \approx 157.97$$)

The numbers from $$20$$ through $$29$$ are all $$\geq20$$, the numbers from $$30$$ through $$39$$ are all $$\geq30$$, and so on, so $$100!>20^{10}\cdot30^{10}\cdot\cdots\cdot90^{10}=(2\cdot3\cdot\cdots\cdot9)^{10}\cdot10^{80}>100^{10}\cdot10^{80}=10^{100}$$

• You need the teens as well: the rest only give you $10^{80}$. Commented Oct 29, 2019 at 17:48
• @NickD Aargh. I seem to be losing the ability to count! Commented Oct 29, 2019 at 21:36
• Welcome to the club! And not that it mattters but the last $\gt$ should be $=$ :-) Commented Oct 29, 2019 at 22:22