3. On the left half of the board, the number 21 is written, and on the right half, the number 15 is written. It is allowed to take an arbitrary number $a$ from the left half of the board and an arbitrary number $b$ from the right half of the board, compute the numbers $a b, a^{3}+b^{3}$, and write $a b$ on the left and $a^{3}+b^{3}$ on the right side of the board, respectively. Can the number 2013201420152016 be obtained on the board using such operations?
# Answer: No  First solution.

Notice that the number $21 \vdots 7$, while 15 gives a remainder of 1. This means that after the first operation, the number $a b \vdots 7$, and the number $a^{3}+b^{3}$ gives a remainder of 1 when divided by 7. Thus, we get that on the left board we will write a number divisible by 7, and on the right board a number giving a remainder of 1 when divided by 7. Consequently, after several such operations, on the left board there will be numbers divisible by 7, and on the right board numbers giving a remainder of 1, but the number 2013201420152016 gives a remainder of 3 when divided by 7.

## Second solution.

Notice that the number $21 \vdots 3$, while $15 \vdots$. This means that after the first operation, the number $a b: 9$, and the number $a^{3}+b^{3}: 9$. Thus, we get that on both the left and right boards we will write a number divisible by 9. Consequently, after several such operations, on the boards there will be numbers divisible by 9 (except for 21 and 15), but the number 2013201420152016 is not divisible by 9.