The functions $ f $ and $ g $ are defined by $ f (x) = 3 ^ x $ and $ g (x) = 100 ^ x $. Two sequences $ a_1, a_2, a_3, \ldots$ and $ b_1, b_2, b_3, \ldots $ are then defined as follows:

(i) $ a_1 = 3 $ and $ a_ {n + 1} = f (a_n) $ for $ n \geq 1 $.

(ii) $ b_1 = $ 100 and $ b_ {n + 1} = g (b_n) $ for $ n \geq $ 1.

Determine the smallest positive integer $ m $ for which $ b_m> a_ {100}$.

$a_n$ is a power tower of $n$ threes and $b_n$ is a power tower of $n$ hundreds. I have read that the first thing that matters in power towers is the height, then the top number matters much more than anything below. We can see $b_{99}>a_{100}$, as we can evaluate the upper $3^3$ on the stack to be $27$, so that each “partial stack” in $b_{99}$ is greater than the corresponding term in the $a_{100}$ stack. To compare $b_{98}$ with $a_{100}$ we can again evaluate the top $3^{3^3}=3^{27}=7625597484987$ to get two power towers with the same number of layers. Since this number is so much greater than $100$, it must be that $a_{100}>b_{98}$ but I don't have a proof.

Here’s an attempt. Let $c = \frac{\log 100}{\log 3} \approx 4.19.$ Define $r_{n,k} = a_{n+k}/b_n$, so that we want $r_{98, 2} > 1.$ Take logs on both sides of $a_{100} > b_{98}$ to get $a_{99} \log 3 > b_{97} \log 100,$ or $r_{97,2} > c.$

  • 2
    $\begingroup$ This sounds like a very hard problem to establish. It might help to read the following thread. $\endgroup$
    – ViHdzP
    Dec 3, 2019 at 1:45
  • $\begingroup$ I unintentionally edited the question $\endgroup$ Dec 3, 2019 at 16:07

1 Answer 1


It seems to me that the last edit basically solves the problem, but here it goes.

To prove $a_{100} > b_{98}$ take $\log_3$ of both sides, to reach the equivalent statement $a_{99} > \log_3(100)b_{97}$. This will be proven if we can prove $a_{99} > 7b_{97}$, since $\log_3(100) < 7$. The choice of the constant $7$ will be clarified soon.

Again, take $\log_3$, to see that you need to prove $a_{98} > \log_3(7) + \log_3(100)b_{96}$. It will be enough to prove $a_{98} > 2 + 5b_{96}$, and for this, it will be enough to prove $a_{98} > 7b_{96}$, since $b_{96} > 1$.

Now continue inductively, until you reach the sufficient statement $a_3 > 7b_1$, which you proved in the original post. As a consequence, $a_{100}>b_{98}$. Since you had already proved $a_{100}<b_{99}$, the least $m$ satisfying the conditions of your problem is $\boxed{m=99}$.

  • $\begingroup$ So what is the solution? $\endgroup$ Dec 3, 2019 at 16:22
  • $\begingroup$ @MeuluElisson It’s $99$. I’ll add that in an edit, for clarity. $\endgroup$
    – ViHdzP
    Dec 3, 2019 at 16:28

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