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Difference between revisions of "1970 IMO Problems/Problem 2"

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Latest revision as of 16:12, 17 November 2024

Problem

Let $a, b$, and $n$ be integers greater than 1, and let $a$ and $b$ be the bases of two number systems. $A_{n-1}$ and $A_{n}$ are numbers in the system with base $a$ and $B_{n-1}$ and $B_{n}$ are numbers in the system with base $b$; these are related as follows:

$A_{n} = x_{n}x_{n-1}\cdots x_{0}, A_{n-1} = x_{n-1}x_{n-2}\cdots x_{0}$,

$B_{n} = x_{n}x_{n-1}\cdots x_{0}, B_{n-1} = x_{n-1}x_{n-2}\cdots x_{0}$,

$x_{n} \neq 0, x_{n-1} \neq 0$.

Prove:

$\frac{A_{n-1}}{A_{n}} < \frac{B_{n-1}}{B_{n}}$ if and only if $a > b$.

Solution

Suppose $a>b$. Then for all integers $0 \le k \le n$, $x_n x_k a^n b^k \ge x_n x_k b^n a^k$, with equality only when $k=n$ or $x_k = 0$. (In particular, we have strict inequality for $k=n-1$.) In summation, this becomes \[x_n a^n \sum_{k=0}^n x_k b^k > x_n b^n \sum_{k=0}^n x_k a^k,\] or \[x_n a^n \cdot B_n > x_n b^n \cdot A_n,\] which is equivalent to \[\frac{x_n a^n}{A_n} > \frac{x_n b^n}{B_n} .\] This implies \[\frac{A_{n-1}}{A_n} = 1 - \frac{x_n a^n}{A_n} < 1 - \frac{x_n b^n}{B_n} = \frac{B_{n-1}}{B_n} .\] On the other hand, if $a=b$, then evidently $A_{n-1}/A_n = B_{n-1}/B_n$, and if $a < b$, then by what we have just shown, $A_{n-1}/A_n > B_{n-1}/B_n$. Hence $A_{n-1}/A_n < B_{n-1}/B_n$ if and only if $a>b$, as desired. $\blacksquare$


Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

Resources

1970 IMO (Problems) • Resources
Preceded by
Problem 1 		1 • 2 • 3 • 4 • 5 • 6 Followed by
Problem 3
All IMO Problems and Solutions
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