sum of $9$ consecutive natural numbers and $10$ consecutive natural numbers are equal 
Find the smallest natural number which can be expressed as sum of $9$
consecutive natural numbers and also $10$ consecutive natural numbers.

I considered these sequences of natural numbers: $x+1,\cdots,x+9\quad\text{and}\quad y+1,y+2,\cdots,y+10\quad$ therfore :
$$9x+45=10y+55=a\quad\text{where}\quad x,y,a\in\mathbb{N}$$
But I don't know how to find minimum of $a$.
 A: Let those numbers be $$x+1,x+2,...,x+9$$ and $$y,y+1,...,y+9$$ Then, $$a = 9x + 45 = 10y + 45 \\ 9x = 10y \implies x = \frac{10y}9$$
Since $\gcd(9,10)=1$, it follows that $9$ divides $y$. Minimum such natural $y$ is $9$. Hence $$a=10\cdot 9 + 45 = 135$$
A: Note that $9x+45 = 9(x+5)$ is a multiple of $9$ at least as large as $54$, and $10y + 55 = 5(2y+11)$ is a multiple of $5$ but not $10$ (why?) and at least as large as $55$.
The smallest number fitting both these criteria must be the multiple of $9$ ending with $5$. Since $9$ and $5$ are co-prime, these are just odd multiples of $45$ (so they don't end with a $0$). Then we just have $45 \times 3 = 135$ as the smallest number fitting the criteria.
Indeed, $135 = 11 + 12 + ... +18+19$ and $135 = 9+10+... + 17+18$.
A: $9x=10y+10\implies y+1\equiv0\mod 9$ so $y=9k-1,k\in\Bbb Z^+$. The smallest such $y$ is $9-1=8$ for which we get $x=10$ and $a=135$.
Note that if we take $0\in\Bbb N$, then the smallest permissible $y$ is $9(0)-1=-1$ for which we get $x=0$ and $a=45$.
A: I guess we’re not including zero as a natural number here or the answer would be very simple (45).
So the sum of a sequence of consecutive natural numbers is average times count. We know the count and the average will be an integer for odd-length sequences and half more than an integer for even length sequences - we can bring that back to an integer by doubling the average to an odd number and halving the count, but the result for a 10-length sequence will be odd
Thus we know that our answer is divisible by $9$ and by $5$. Discarding $45$ the next is $90$, which is not odd so we move on to $135$. This implies that the average is $15$ for the length-9 sequence and $13.5$ for the length-10. So
$$ 11+12+13+14+15+16+17+18+19 = 8+9+10+11+12+13+14+15+16+17 =135$$
