Alex Chui didn’t just win the International Mathematical Olympiad last month. He broke it. The British student, a year 13 pupil at Tonbridge School in Kent, walked away with a perfect score. That’s 100 percent. It secured him the gold medal and raised his total medal count to seven—five of them gold.
This is historic. No contestant in the IMO’s long history since 1959 has ever won a medal for seven consecutive years. The competition pulls together school-age talent from over a hundred countries. Usually, you can’t sustain that level of precision for nearly a decade. Chui changed the rules of the game in Shanghai.
But the math doesn’t stop at medals. I asked Chui what puzzle he liked best. He didn’t give me something from his exam. He offered this instead.
The 3×3 grid challenge
Here is the setup. You need to fill a 3×3 grid.
The cells must contain positive whole numbers.
There is a constraint. The product of the numbers in each row must equal 30.
The product of the numbers in each column must also equal 30.
Simple? No. If you try to count these grids by brute force, you will get lost. There are more than 200 valid ways to fill this grid. Trial and error is a trap. The numbers don’t have to be unique. Repetition is allowed.
The answer isn’t about guessing. It’s about breaking down 30 into its prime components.
So, how do you solve it without drowning in 200 possibilities? You look at the factors.
Prime factorization is the key
30 is a small number, but it’s composed of specific building blocks.
30 = 2 × 3 × 5
Those are primes. There are no other integer factors that matter here. Any number you place in the grid must be formed by combinations of these three primes. Since the product of three numbers in a row is 30, each row must distribute these primes.
Let’s look at the diagonal. Or just the first row.
In any single row, you are multiplying three numbers. The result is 2 × 3 × 5.
This means each row contains the prime factors 2, 3, and 5, scattered across three cells. Some cells might just hold 1 (which is neither 2, 3, nor 5). Others might hold a composite like 6 (2×3) or 10 (2×5) or 15 (3×5). But the total count of 2s, 3s, and 5s across the row must match the target.
Breaking the symmetry
Here is where it gets elegant.
Consider the diagonal from top-left to bottom-right. Let the grid be $G$.
$G_{11} \times G_{22} \times G_{33} = \text{something?}$ Not necessarily 30. That’s not the rule.
The rule is row products and column
