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, 2020

Algebra Difficulty 6.5 National Olympiad Prove it United States

Problem:

The classrooms at MIT are each identified with a positive integer (with no leading zeroes). One day, as President Reif walks down the Infinite Corridor, he notices that a digit zero on a room sign has fallen off. Let $N$ be the original number of the room, and let $M$ be the room number as shown on the sign.

The smallest interval containing all possible values of $\frac{M}{N}$ can be expressed as $\left[\frac{a}{b}, \frac{c}{d}\right)$ where $a, b, c, d$ are positive integers with $\operatorname{gcd}(a, b)=\operatorname{gcd}(c, d)=1$ . Compute $1000 a+100 b+10 c+d$ .

Solution

Solution:

Let $A$ represent the portion of $N$ to the right of the deleted zero, and $B$ represent the rest of $N$ . For example, if the unique zero in $N=12034$ is removed, then $A=34$ and $B=12000$ . Then, $\frac{M}{N}=\frac{A+B / 10}{A+B}=1-\frac{9}{10} \frac{B}{N}$ .

The maximum value for $B / N$ is 1, which is achieved when $A=0$ . Also, if the 0 removed is in the $10^{k}$ 's place ( $k=2$ in the example above), we find that $A<10^{k}$ and $B \geq 10^{k+1}$ , meaning that $A / B<1 / 10$ and thus $B / N>10 / 11$ . Also, $B / N$ can get arbitrarily close to $10 / 11$ via a number like $1099\ldots 9$ .

Therefore the fraction $\frac{M}{N}$ achieves a minimum at $\frac{1}{10}$ and always stays below $\frac{2}{11}$ , though it can get arbitrarily close. The desired interval is then $\left[\frac{1}{10}, \frac{2}{11}\right)$ .

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.