2. How many three-digit natural numbers are there such that the digit in the hundreds place is equal to the sum of the digits in the tens and units places?
Solution. Three-digit numbers that satisfy the condition of the problem are of the form $\overline{a b c}, a \neq 0$ and $a=b+c$.

- for $a=1$ these are the numbers: 101,110,
- for $a=2$ these are the numbers: 202, 211, 220,
- for $a=3$ these are the numbers: $303,312,321,330$,
- for $a=4$ these are the numbers: 404, 413, 422, 431, 440,
- for $a=5$ these are the numbers: $505,514,523,532,541,550$,
- for $a=6$ these are the numbers: $606,615,624,633,642,651,660$,
- for $a=7$ these are the numbers: $707,716,725,734,743,752,761,770$,
- for $a=8$ these are the numbers: $808,817,826,835,844,853,862,871,880$,
- for $a=9$ these are the numbers: 909, 918, 927, 936, 945, 954, 963, 972, 981, 990.

Finally, there are $2+3+4+5+6+7+8+9+10=54$ three-digit numbers with the desired property.