Which of the following numbers is greater? Without using a calculator and logarithm.

$$100^{101} , 101^{100}$$

My try : $$100=10^2\\101=(100+1)=(10^2+1)$$

So :


Now what ?

  • 3
    $\begingroup$ Hint: the function $x^{1/x}$ is decreasing for $x>e$. $\endgroup$ – lulu Nov 27 '17 at 19:43
  • $\begingroup$ A special case of math.stackexchange.com/questions/517555/… $\endgroup$ – Henry Nov 27 '17 at 23:34
  • $\begingroup$ @lulu: as is $\dfrac{\log x}{x}$ $\endgroup$ – Henry Nov 27 '17 at 23:36
  • 1
    $\begingroup$ @Henry right, same thing. $\endgroup$ – lulu Nov 27 '17 at 23:44
  • $\begingroup$ These boundary methods are usually employed in Computing Science. $\endgroup$ – Dehbop Nov 28 '17 at 5:00

You want to determine if $\left(\frac{101}{100}\right)^{100}\geq 100$. But we know that $ \left(1+\frac{1}{n}\right)^n$ is always less than $e$.

  • $\begingroup$ This is the nicest solution. I love it $\endgroup$ – Sopel Nov 28 '17 at 9:47

101100 = (100+1)100

Using the binomial theorem, we get 101 terms, none of them greater than 100100, and the three smallest of these terms add up to less than 100100. So the sum is less than 99(100100), which is less than 100(100100), which is equal to 100101.

  • 7
    $\begingroup$ This the only answer that doesn't require knowledge of $e$. $\endgroup$ – MackTuesday Nov 28 '17 at 6:33

$$\frac{101^{100}}{100^{101}} =\frac1{100}\cdot \left(\frac{101}{100}\right)^{100}=\frac1{100}\cdot\left(1+\frac1{100}\right)^{100}\approx \frac e{100}\ll 1$$

Of course, the "$\approx$" needs to be tamed, but the "$\ll$" allows for much leeway. One way to elaborate the idea is to use $$ \left(1+\frac1{100}\right)^{100}\cdot \left(1-\frac1{100}\right)^{100}=\left(1-\frac1{10000}\right)^{100}<1$$ and Bernoulli's inequality: $$\left(1-\frac1{100}\right)^{100}=\left(1-\frac1{100}\right)^{50}\cdot \left(1-\frac1{100}\right)^{50}\ge \left(1-\frac{50}{100}\right)^2=\frac14$$ so that ultimately $$\frac{101^{100}}{100^{101}}<\frac 4{100}. $$

  • 2
    $\begingroup$ Didn't you meant $\frac{1}{100}$ as first term after first equality? $\endgroup$ – Zereges Nov 27 '17 at 20:16
  • $\begingroup$ About taming the approximation, doesn't $(1+1/n)^n = e'$ approach $e$ from below, so that $e' < e$ always? And then $e'/100 < 1$ $\endgroup$ – ilkkachu Nov 28 '17 at 9:04
  • $\begingroup$ I don't see what good is multiplying $\left(1+\frac1{100}\right)^{100}$ with $\left(1-\frac1{100}\right)^{100}$, arrive at $\left(1-\frac1{10000}\right)^{100}$??? $\endgroup$ – Dehbop Nov 28 '17 at 10:06

In this answer, it is shown that $\left(1+\frac1n\right)^n$ increases to $e\lt3$. Thus, for $n\ge3$, $$ \left(1+\frac1n\right)^n\lt n $$ Multiply by $n^n$ and we get $$ (n+1)^n\lt n^{n+1} $$


for a sequence, $100^n$, as n increases, the number of digits increases by 2.

In the sequence of $101^n$, after finding the first 3 terms one can see the pattern and conclude that for all values of n, the term is 2 digits more than the last term, never 3 digits more. $101, 10201, 1030301, 104040401, 10505050501\dotsc$

If we compare $100^{101}$ and $101^{100}$, we can use the two above observations to conclude that $100^n > 101^{n-1}$, so $100^{101} > 101^{100}$.

  • $\begingroup$ I'm not convinced by your argument about the number of digits in $101^n$. Consider, for example, $101^{231}$ vs $101^{232}$. Indeed, we know that your argument eventually fails, because $101^n \gg 100^n$ when $n$ is very large. $\endgroup$ – Xander Henderson Nov 28 '17 at 3:44
  • $\begingroup$ 101^n is always larger than 100^n unless n is 0, so what you said is correct but doesn't dispute what I stated, ill keep looking at larger scales though $\endgroup$ – user507719 Nov 28 '17 at 3:47
  • $\begingroup$ the sequence can be written using recursion as A(n)=A(n-1)+A(n-1)*100. in the case you mentioned, 101^232 would be equal to 101^231+101^231*100. These differ by 2 units. This means that 101^231<100^232 (sorry I'm new to using a computer to write math so I don't know how to make it display the correct notation) $\endgroup$ – user507719 Nov 28 '17 at 4:02
  • $\begingroup$ looking at the sequence of $101^n$ what I see is that it starts with 100+n as the first digits. This suggests to me that if n is over 1000 that it will add more digits. That n is bigger than this current case but it means that your "never 3 digits more" is provably wrong and thus needs to be better specifed. In fact I believe this can be seen more easily by considering that the number of digits in a number is one more than the integer part of the base 10 log of that number. In this case that would be $n*log(101)$ which goes up by three from 231 to 232 (as Xander implied above). $\endgroup$ – Chris Nov 28 '17 at 10:04

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