There are many identical squares. In the vertices of each of them, the numbers 1, 2, 3, and 4 are written in random order. The squares are stacked, and the sum of the numbers that ended up in each of the four corners of the stack is written down. Can it be that:

a) in each corner of the stack, the sum is 2004?

b) in each corner of the stack, the sum is $2005?$
a) The sum of all numbers should be divisible by $1+2+3+4=10$. But 2004$\cdot$4 is not divisible by 10.

b) We can take 401 pairs of squares with such a number arrangement $(1,2,3,4)$ and ( $4,3,2,1$ ).

## Answer

a) It cannot; b) it can.