Argentina_2018

Question

The rows and the columns of a 16 16 table are labeled from 1, 2, , 16, and the product i j is written in the square in row i, column j. Several rows are chosen (at least 2) and also several columns (at least 2). Then the numbers at their intersections are deleted. a) Can the sum of all deleted numbers be a prime? b) What about the sum of all undeleted numbers?

Accepted Answer

a) The sum of the deleted numbers is always composite. Let the chosen rows be

i_{1}, \dots, i_{p}

,

p \geq 2

and the chosen columns

j_{1}, \dots, j_{q}

,

q \geq 2

. The deleted numbers in row

i_{1}

are

i_{1} j_{1}, i_{1} j_{2}, \dots, i_{1} j_{q}

. Likewise the deleted numbers in row

i_{2}

are

i_{2} j_{1}, i_{2} j_{2}, \dots, i_{2} j_{q}

and so on; for row

i_{p}

they are

i_{p} j_{1}, i_{p} j_{2}, \dots, i_{p} j_{q}

. The total deleted sum is therefore

S = (i_{1} + \dots + i_{p}) (j_{1} + \dots + j_{q})

. Both factors are at least

2

(as

p \geq 2, q \geq 2

), hence

S

is composite.

b) The sum of all undeleted numbers can be a prime. Let the chosen rows be

2, 3, \dots, 16

and the chosen columns

2, 3, \dots, 16

. Then the undeleted numbers are in the union of row

1

and column

1

. Their sum is

2 (1 + 2 + \dots + 16) - 1 = 271

, which is a prime.