Which is greater: $101^{99}\text{ or }100^{100}$?

Question

So I was scrolling through the homepage of Youtube when I came across this math problem by Learn with Christian Expo which proposed the following question:$$\text{Which is greater: }101^{99}\text{ or }100^{100}\text{?}$$Here is my attempt at solving the proposed question (both attempts):

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$$\text{Attempt }1$$

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My first attempt was utterly stupid (probably along with my second attempt, since I honestly do not know how to solve these types of problems). I decided that, knowing already that $100^{100}$ was just a $1$ followed by $200$ zeros, decided to attempt to use Pascal's Triangle to calculate the powers of $101$

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$$101^1=101$$$$101^2=10201$$$$101^3=1030301$$$$101^4=104060401$$$$101^5=10510100501$$$$101^6=1061520150601$$$$101^7=107213535210701$$$$101^8=10828567056280801$$Now things get messy after powers of $8$. Here's the same exact triangle, just that we aren't simplifying any numbers:$$101^1=101$$$$101^2=10201$$$$101^3=1030301$$$$101^4=104060401$$$$101^5=1050(10)0(10)0501$$$$101^6=1060(15)0(20)0(15)0601$$$$101^7=1070(21)0(35)0(35)0(21)0701$$$$101^8=1080(28)0(56)0(70)0(56)0(28)0801$$$$101^9=1090(36)0(84)0(\color{red}{126})0(\color{red}{126})0(84)0(36)0901$$Which simplifying the numbers gets us$$101^9=1093685272684360901$$$$101^{10}=110462212541120451001$$Obviously, calculating this through Pascal's Triangle is very tedious, so I decided to call it quits on the first attempt.

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$$2\text{nd attempt}$$

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Knowing that $d^c=(ab)^c=a^c\cdot b^c$ where $ab=d$, and knowing that $100^{100}$ can be rewritten as $2^{100}\cdot2^{100}\cdot5^{100}\cdot5^{100}$, and also knowing that$$2^{100}\cdot2^{100}=1267650600228229401496703205316^2$$And that$$5^{100}\cdot5^{100}=7888609052210118054117285652827862296732064351090230047702789306640625^2$$I can see how many times these will divide into multiples of $101$ before getting too large. Here is my attempt at doing that:$$101^1\gt\lfloor2^n\rfloor\text{, }-\infty\leq n\lt8$$$$101^2\gt\lfloor2^n\rfloor\text{, }-\infty\leq n\lt14$$$$101^3\gt\lfloor2^n\rfloor{, }-\infty\leq n\lt20$$$$\therefore101^{33}\gt\lfloor2^n\rfloor\text{, }-\infty\leq n\lt200$$$$\therefore\frac{101^{99}}{2^{200}\cdot5^{200}}\approx\frac{101^{66}}{5^{200}}$$$$101^1\gt\lfloor5^n\rfloor\text{, }-\infty\leq n\lt2$$$$101^2\gt\lfloor5^n\rfloor\text{, }-\infty\leq n\lt6$$$$101^3\gt\lfloor5^n\rfloor\text{, }-\infty\leq n\lt9$$$$101^4\gt\lfloor5^n\rfloor\text{, }-\infty\leq n\lt12$$$$101^5\gt\lfloor5^n\rfloor\text{, }-\infty\leq n\lt15$$$$\therefore101^{66}\gt\lfloor5^n\rfloor\text{, }-\infty\leq n\lt61\frac{2}{3}$$$$\therefore0\lt\frac{100^{66}}{5^{200}}\lt1$$$$\therefore101^{99}\lt100^{100}$$

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$$\mathbf{\text{My question}}$$

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Would at least my second attempt on this problem be correct, or if it is not, what could I do to attain the correct solution more easily?

Answer 1

Here is an elementary proof without calculator. The inequality $$ 101^{99} < 100^{100} $$ is equivalent to $$ \left( 1 + \frac{1}{100} \right)^{100} < 101 \, . $$
One then proves by induction that for all $n \ge 2$ $$ \left( 1 + \frac{1}{n}\right)^n < n+1 $$ which implies the desired inequality for $n = 100$.

The base case $n = 2$ is clearly true.

Assume the statement is true for some $n \ge 2$, then $$ \begin{align} \left( 1 + \frac{1}{n+1}\right)^{n+1} &= \left( 1 + \frac{1}{n+1}\right)^n \cdot \frac{n+2}{n+1}\\ &< \left( 1 + \frac{1}{n}\right)^n \cdot \frac{n+2}{n+1}\\ &< (n+1) \cdot \frac{n+2}{n+1} = n+2 \, . \end{align} $$

Answer 2

Here's another way using Bernoulli's Inequality:

$$(1+n)^{(n-1)/n} \leqslant 1+n\times\frac{n-1}{n}=n$$ $$\implies (n+1)^{(n-1)}\leqslant n^n$$

Equality requires $n=1$ so for $n=100$ this is strict.

Answer 3

Well, I think your second solution does head in a reasonable direction, albeit, a little tedious. Here is what I would do:

While we do not know if $101^{99}$ is greater or less than $100^{100}$, we know that $101^{99} = 100^x$ for some value of $x$. So we solve. If $x>100$, then $101^{99}>100^{100}$ and if $x<100$, then we have the opposite, $101^{99}<100^{100}$.

So, using logarithm properties (use natural log or log base 10 - I prefer $\ln x$)

$$ 101^{99} = 100^x $$ $$ 99 \ln (101) = x \ln (100)$$ $$x = 99 \frac{\ln(101)}{\ln (100)}$$ $$ x \approx 99.2139$$

Thus, we see that $101^{99} \approx 100^{99.2139}$ and we would get a larger number on the right hand side if we increase the exponent on the base $100$. Hence, $$ 101^{99} < 100^{100}.$$

Answer 4

We have that for $x>1$

$$f(x)=x\log x -(x-1)\log(x+1) \implies f'(x)=\frac2{x+1}-\log \left(1+\frac1x\right)>\frac2{x+1}-\frac1x>0$$

therefore $f(x)>0$ and by $x=100$

$$100\log 100 -99\log 101 >0 \iff \log 100^{100}>\log 101^{99} \iff 100^{100}> 101^{99}$$

Answer 5

As well as induction and logarithms, you could use the binomial expansion: $$101^{99}=(100+1)^{99}=100^{99}+\binom{99}{1}100^{98}+\binom{99}{2}100^{97}+\cdots+\binom{99}{r}100^{99-r}+\cdots+99\cdot100+1$$ There are $100$ terms on the right-hand side, and each of them except the first is less than $100^{99}$. So their sum is less than $100\cdot 100^{99}=100^{100}$.

To see that each term except the first is less than $100^{99}$:

$$\begin{align} \binom{99}{r}100^{99-r}&=\frac{99!}{r!(99-r)!}100^{99-r}\\ &<\frac{99^r}{r!}100^{99-r}\,\,\,\,\,\,\,\,\text{if $r>0$}\\ &\le 99^r\cdot100^{99-r}\\ &=100^{99} \end{align}$$