Problem Log - 2023 IOQM No. 3

by Nibir Sankar, Jun 15, 2025

Problem

Let $α$ and $β$ be positive integers such that

\frac{16}{37} < \frac{α}{β} < \frac{7}{16}

Find the smallest possible value of β .

Reciprocal the inequality:

\begin{array}{l} \frac{16}{7} < \frac{β}{α} < \frac{37}{16} \end{array}

(the terms will switch sides)

Now, multiply all the terms with $α$ :

\begin{array}{l} \frac{16}{7} α < β < \frac{37}{16} α \end{array}

Now, $\frac{16}{7} \approx 2. 28$ and $\frac{37}{16} \approx 2.31$

We need to pick an $α$ such that $β$ has a solution in the $Z$ and that too, the smallest.

Through trial and error, we may find that if $α = 10$ , then

\begin{array}{l} 22.8 < β < 23.1 \\ ∴ β = 23 \end{array}

Which is the smallest possible value.

What tripped me?
- The fact that it took me over 10 minutes is a disgrace. (It took 47m). But since I am a learning mathematician, that is normal. I learned a lot from the problem.
What pattern did I recognize?
- I saw that the differences between the inequalities (no matter what $α$ you pick) remains almost negligible.
- I tried $α = L C M (7, 16)$ , but that was too large.
What did I learn?
- I was trying HUGE numbers for $α$ , when I could’ve tried $α = 10$ .
- It took me 47m to get there,
- But the next time, whenever there will be decimals, or inequalities, I will try smaller numbers and 10.

🧠: 1–5 for [Insight:: 4] ⚔️: 1–5 for [Difficulty:: 2]