Galperin G.A.

For each pair of numbers $x$ and $y$, a certain number $x^{*} y$ is assigned. Find 1993*1935, given that for any three numbers $x, y, z$ the identities $x^{*} x=0$ and $x^{*}\left(y^{*} z\right)=\left(x^{*} y\right)+z$ hold.
$x=x^{*} x+x=x^{*}\left(x^{*} x\right)=x^{*} 0=x^{*}\left(y^{*} y\right)=x^{*} y+y$. Therefore, $x^{*} y=x-y$. Hence $1993 * 1935=1993-1935=58$.

## Answer

58.