2. The sum of 16 consecutive natural numbers, including the number $n$, is fifteen times greater than the number $n$. Determine all numbers $n$ for which this is possible.
## First Solution.

Let the smallest of 16 consecutive numbers be denoted by $x$. Then we have:

$$
x+(x+1)+(x+2)+\cdots+(x+15)=15(x+k)
$$

where $n=x+k$ for some $k \in\{0,1, \ldots, 15\}$. Since $1+2+\cdots+15=120$, by rearranging we get $16 x+120=15 x+15 k$ or $x=15 k-120=15(k-8)$.

For $x$ to be a natural number, the difference $k-8$ must be a positive integer, which means $k>8$.

Let's examine all possibilities:

for $k=9$ we have $x=15(9-8)=15, n=x+k=24$, and $15+16+\ldots+30=15 \cdot 24$

for $k=10$ we have $x=15(10-8)=30, n=x+k=40$, and $30+31+\ldots+45=15 \cdot 40$

for $k=11$ we have $x=15(11-8)=45, n=x+k=56$, and $45+46+\ldots+60=15 \cdot 56$

for $k=12$ we have $x=15(12-8)=60, n=x+k=72$, and $60+61+\ldots+75=15 \cdot 72$

for $k=13$ we have $x=15(13-8)=75, n=x+k=88$, and $75+76+\ldots+90=15 \cdot 88$

for $k=14$ we have $x=15(14-8)=90, n=x+k=104$, and $90+91+\ldots+105=15 \cdot 104$

for $k=15$ we have $x=15(15-8)=105, n=x+k=120$, and $105+106+\ldots+120=15 \cdot 120$

The sought numbers are 24, 40, 56, 72, 88, 104, and 120.