## Task B-1.6.

The sum of the digits of a three-digit number is 17. The digit in the tens place is 9. If the digits in the hundreds and units places are swapped, the new number is 100 more than twice the original number. What is the original number?
## First solution.

We write the required number in the form $\overline{a 9 b}$. It holds that $a+9+b=17, a+b=8$. 2 points

The new number is $\overline{b 9 a}$, and according to the condition of the problem, $\overline{b 9 a}=100+2 \cdot \overline{a 9 b}$. 1 point

Thus, $100 b+90+a=100+2(100 a+90+b)$. 1 point

It follows that $100 b-2 b=200 a-a+280-90, 98 b=199 a+190$. 1 point

Since $a+b=8, b=8-a$, it follows that

\[
\begin{array}{r}
98(8-a)=199 a+190 \\
784-98 a=199 a+190 \\
-297 a=-594 /:(-297)
\end{array}
\]

Thus $a=2$, and $b=8-a=6$.

The required number is $\overline{a 9 b}=296$.