74th NMO Selection Tests for JBMO

a) Let $p_1,p_2,...,p_7$ be the products of the elements of the rows $1,2,...,7$. We have $p_1=p_7$, $p_2=p_6$ and $p_3=p_5$, so $49!=(p_1{p}_2{p}_3{)}^2\cdot p_4$, (1).

Note that $49!=2^{46}\cdot 3^{22}\cdot 5^{10}\cdot 7^8\cdot 11^4\cdot 13^3\cdot 17^2\cdot 19^2\cdot 23^2\cdot 29\cdot 31\cdot 37\cdot 41\cdot 43\cdot 47$. From (1) we infer that the primes which have odd exponents in the prime factorization of $49!$ must appear in the prime factorization of one of the numbers situated on the $4$-th row, thus $13\cdot 29\cdot 31\cdot 37\cdot 41\cdot 43\cdot 47\mid p_4$. The product of any two of the primes $13,29,31,37,41,43$ and $47$ exceeds $49$, therefore each of these primes divides exactly one of the $7$ numbers situated on the $4$-th row.

The number $\frac{49!}{p_4}$ is a perfect square, hence if $p_4>13\cdot 29\cdot 31\cdot 37\cdot 41\cdot 43\cdot 47$, we have that $p_4\ge 2^2\cdot 13\cdot 29\cdot 31\cdot 37\cdot 41\cdot 43\cdot 47$, thus one of the numbers written in the cells of the table will be at least equal to $\min \{2^2\cdot 13,2\cdot 29\}>49$, false.

Therefore, the $4$-th row contains the numbers $13,29,31,37,41,43,47$, whose sum is the prime $241$.