For each positive integer $n$, let $I_n$ denote the number of integers $p$ for which $50^n<7^p<50^{n+1}$. (a) Prove that, for each $n$, $I_n$ is either $2$ or $3$. (b) Prove that $I_n=3$ for infinitely many $n\in\mathbb N$, and find at least one such $n$.
Problem
Source: France 1994 P1
Tags: Sequence, algebra
BarisKoyuncu
07.05.2021 20:31
I think there is a typing error. $7^p<50^{p+1}$ doesn't make any sense. It can be $50^n<7^p<50^{n+1}$
jasperE3
07.05.2021 20:33
Yes, the version I used had an error, it should be fixed now.
BarisKoyuncu
07.05.2021 20:44
Suppose $7^{p-1}<50^n<7^p$ for a given $n$.
First, let's show that $I_n<4$.
For all integers $t$ greater than $p+2$, we have $7^t\ge 7^{p+3}=343\cdot 7^p>50\cdot 7^p>50^{n+1}$. Thus, $I_n<4$.
Now, let's show that $I_n\ge 2$.
We know $p$ satisfies the condition. Also, $7^{p+1}>7^p>50^n$ and $7^{p+1}=7^{p-1}\cdot 49<7^{p-1}\cdot 50<50^n\cdot 50=50^{n+1}$. Thus, $I_n\ge 2$.
cadaeibf
07.05.2021 22:27
For part a, note that the inequality is equivalent to $n\log_7(50)<p<(n+1)\log_7(50)$, and since $2<\log_7(50)<3$ we can only have two or three integers between $n\log_7(50)$ and $(n+1)\log_7(50)$. For part b, if we want three solutions we have to have $\{(n+1)\log_7(50)\}<\log_7(50)-2$, but there exist infinitely many such $n$ by Kronecker density theorem, and so we are done. Also, through some calculations, $n=96$ works (or also $0$ if you include it in $\mathbb{N}$).