B1. Find all real numbers $x$ for which $\left(x^{2}-7 x+11\right)^{x^{2}-13 x+42}=1$.
B1. The value of the power $a^{b}$ is equal to 1 only in the case when $a=1$, $a=-1$ and $b$ is an even integer, or $b=0$ and $a \neq 0$. If $x^{2}-7 x+11=1$, it follows that $x^{2}-7 x+10=0$ or $(x-2)(x-5)=0$. From this, we get the solutions $x=2$ and $x=5$. If $x^{2}-7 x+11=-1$ or $x^{2}-7 x+12=0$, it follows that $(x-3)(x-4)=0$ and thus $x=3$ or $x=4$. In both cases, $x^{2}-13 x+42$ is an even number, so these are also solutions. If, however, $x^{2}-13 x+42=0$ or $(x-6)(x-7)=0$, it follows that $x=6$ or $x=7$ and in both cases $x^{2}-7 x+11 \neq 0$. The solutions are therefore the numbers $2,3,4,5,6$ and 7. [^0]