FINAL ROUND

Question

Real numbers are written in the cells of the 7 7 table so that the product of the numbers in any 3 3 square is equal to the product of the numbers in any 4 4 square. Is it possible for the product of all numbers in the table to be 2015? (V. Kaskevich)

Accepted Answer

Answer: yes, it is possible. Let a and b be real numbers such that ab = 1. Consider the following table | a | b | a | b | a | b | a | |-----|-----|-----|-----|-----|-----|-----| | b | a | b | a | b | a | b | | 1 | 1 | 1 | 1 | 1 | 1 | 1 | | a | b | a | b | a | b | a | | b | a | b | a | b | a | b | | 1 | 1 | 1 | 1 | 1 | 1 | 1 | | a | b | a | b | a | b | a | It is easy to see that the products of the numbers in all 3 3 squares and 4 4 squares are equal to 1. Moreover, the product of all numbers in the table is equal to a, so this product can admit any value different from 0, in particular this product can be equal to 2015.

$a$	$b$	$a$	$b$	$a$	$b$	$a$
$b$	$a$	$b$	$a$	$b$	$a$	$b$
$1$	$1$	$1$	$1$	$1$	$1$	$1$
$a$	$b$	$a$	$b$	$a$	$b$	$a$
$b$	$a$	$b$	$a$	$b$	$a$	$b$
$1$	$1$	$1$	$1$	$1$	$1$	$1$
$a$	$b$	$a$	$b$	$a$	$b$	$a$

$a$	$b$	$a$	$b$	$a$	$b$	$a$
$b$	$a$	$b$	$a$	$b$	$a$	$b$
$1$	$1$	$1$	$1$	$1$	$1$	$1$
$a$	$b$	$a$	$b$	$a$	$b$	$a$
$b$	$a$	$b$	$a$	$b$	$a$	$b$
$1$	$1$	$1$	$1$	$1$	$1$	$1$
$a$	$b$	$a$	$b$	$a$	$b$	$a$

$a$	$b$	$a$	$b$	$a$	$b$	$a$
$b$	$a$	$b$	$a$	$b$	$a$	$b$
$1$	$1$	$1$	$1$	$1$	$1$	$1$
$a$	$b$	$a$	$b$	$a$	$b$	$a$
$b$	$a$	$b$	$a$	$b$	$a$	$b$
$1$	$1$	$1$	$1$	$1$	$1$	$1$
$a$	$b$	$a$	$b$	$a$	$b$	$a$

$a$	$b$	$a$	$b$	$a$	$b$	$a$
$b$	$a$	$b$	$a$	$b$	$a$	$b$
$1$	$1$	$1$	$1$	$1$	$1$	$1$
$a$	$b$	$a$	$b$	$a$	$b$	$a$
$b$	$a$	$b$	$a$	$b$	$a$	$b$
$1$	$1$	$1$	$1$	$1$	$1$	$1$
$a$	$b$	$a$	$b$	$a$	$b$	$a$