74th Romanian Mathematical Olympiad

Since $11^2>100$, the table cannot contain numbers divisible by squares of prime numbers $p$, with $p\ge 7$. Suppose that there exists a number of the table (situated on line $\ell$ and column $c$) divisible by a prime $p\ge 17$. Then there exists one more number on line $\ell$ (and column $c^{\prime }\ne c$) divisible by $p$. Also, there exists one more number on column $c$ (and line ${\ell }^{\prime }\ne \ell$) divisible by $p$. Therefore, in the cell $({\ell }^{\prime },c^{\prime })$ must be a number divisible by $p$. In consequence, in the table must appear a number $N\ge 7p>100$ – impossible.

11	$3\cdot 11$	$3^3$	$5^2$
$5\cdot 11$	$7\cdot 11$	$3\cdot 5$	$3\cdot 7$
$7^2$	$3\cdot 5^2$	13	$3\cdot 13$
$3^2\cdot 5$	$3^2\cdot 7$	$5\cdot 13$	$7\cdot 13$

It follows that the table cannot contain 2-digit numbers that are odd multiples of 17 – 3 numbers, of 19 – 3 numbers, of 23 – 2 numbers, of 29 – 2 numbers, of 31 – 2 numbers and of 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97 – one number each; in total, 26 numbers. On the other hand, there are 45 odd 2-digit numbers, so the table can contain at most $45-26=19$ numbers. Hence $n\le 4$. The table from above is an example for $n=4$.