How Many Lucky Fractions Have Denominator 2025? Olympiad Problem and Solution

A Perfect Brain Warm-Up: Junior Olympiad Problems

The best warm-up for your brain is olympiad problems from the junior grades. Here is one such problem from the Vysshaya Proba competition, contributed by Alexander Shtern:

A fraction with natural numerator and denominator is called lucky if it equals some fraction with natural numerator and denominator in which the numerator is exactly 1 less than the denominator. For example, the fraction 4/6 is lucky. How many lucky fractions with denominator 2025 exist?

Understanding the Lucky Fractions Problem

Let’s unpack what the condition means. A fraction a/2025 is lucky if it equals some fraction of the form (n − 1)/n, where n is a natural number. In other words:

a/2025 = (n − 1)/n

The fraction (n − 1)/n is simply one step below a whole: 1/2, 2/3, 3/4, 4/5, and so on. The example 4/6 = 2/3 fits perfectly — the numerator 2 is one less than the denominator 3.

Setting Up the Equation

From the equation a/2025 = (n − 1)/n, we cross-multiply:

an = 2025(n − 1) = 2025n − 2025

Rearranging:

2025 = n(2025 − a)

So n = 2025 / (2025 − a).

For n to be a natural number, the quantity (2025 − a) must be a positive divisor of 2025. Since a must itself be a natural number (a ≥ 1), we need 2025 − a ≤ 2024, meaning only divisors of 2025 that are strictly less than 2025 are valid.

Counting Divisors: The Key Step in Solving This Olympiad Fraction Problem

Each valid divisor d of 2025 with d < 2025 gives exactly one lucky fraction: a = 2025 − d, and the resulting fraction is a/2025.

So we need to count the divisors of 2025, then subtract 1 (to exclude 2025 itself).

First, factorize 2025:

2025 = 45² = (3² · 5)² = 3⁴ · 5²

The number of divisors is:

τ(2025) = (4 + 1)(2 + 1) = 5 · 3 = 15

Excluding the divisor 2025 itself, we get 15 − 1 = 14 valid divisors.

Answer

There are 14 lucky fractions with denominator 2025.

Each one corresponds to a divisor d of 2025 in the range 1 ≤ d ≤ 1012 (or more precisely, any divisor d < 2025), producing the fraction (2025 − d)/2025, which simplifies to (n − 1)/n for n = 2025/d.