Consider a set $A$ of $10$ integers, $A = \{a_1,a_2, \ldots ,a_{10}\}$. Prove that there is at least one subset of $A$ whose sum is divisible by $10$. Consider a set A of 10 integers, $A= \{a_1,a_2, \ldots ,a_{10}\}$. Prove that there is at least one subset of A whose sum is divisible by 10. Use the pigeon-hole principle.
Hint: Consider the 10 sums $s_1 = a_1, s_2 = a_1 + a_2, \ldots ,s_{10} = a_1 + a_2 + \ldots+a_{10}$. If one of these ten sums is divisible by $10$, we are done. So assume that none of these ten sums is divisible by $10$.
I'm pretty sure this is the proof but im not exactly sure what's $b_l$ and $b_k$ stand for?  
Let $A=\{a_1,a_2,...,a_{10}\}$
Define set of numbers $b_n=\sum\limits_{i=1}^n a_n$ i.e. the set $$\{a_1, (a_1+a_2),(a_1+a_2+a_3),\ldots ,(a_1+a_2+\ldots +a_{10})\}$$
If any element of the set is divisible by $10$, then we are done.
Hence proving case $1$.
Case 2: Assume that no element of this set is disible by $10$.
Then, by Pigeon hole priciple, there exists some $k<l$, such that, $b_l\equiv b_k$ 
then $b_l\equiv b_k (mod 10)$ with $l>k>0$ 
Hence, $\{b_k,b_{k+1}, \ldots,b_l\}$ is a subset of $A$ whose sum is divisble by $10$.
Thus proving Case $2$.
 A: $$b_n = \sum_{i=1}^n a_n.$$
$b_n$ is the sum of the first $n$ $a_i$'s.
In terms of modulo $10$, since none of them is $0 \mod 10$, by pigeonhole principle, we can find a pair of $b_i$, call them $b_l$ and $b_k$ such that 
$$ b_l \equiv b_k \pmod{10}.$$
Hence $b_k - b_l \equiv 0 \pmod{10}$, 
Hence $\sum_{n=k+1}^la_n$ is divisible by $10$.
Hence $\color{red}{\{a_{k+1}, \ldots, a_l\}}$ is one such subset.
I have highlighted a potential fix for a typo in your proof.
A: This is a general proof for:
In a list of $N$ integers, there exists a non-empty subset of this list whose sum is divisible by $N$.
Statement $1$: If $a_1$ and $a_1+a_2+a_3$ have the same remainder when divided by $N$, $a_2+a_3$ must be divisible by $N$.
We define a set $A$ with $N$ elements (this is our list):
$$A = \{a_1,a_2,a_3...a_N\}$$
Then we define $B$ as a set where $B_n = \sum_{i=1}^{n}a_i$
$$B = \{a_1,a_1+a_2,a_1+a_2+a_3...B_N\}$$
When a number is divided by $N$, there are $N$ possible remainders $(0,1,2...N-1)$. We have $N$ elements in $B$. If every remainder is different, one of the remainders must be $0$, therefore, we have a subset whose sum is divisible by $N$. However, if not every remainder is different, then at least two numbers must share a remainder (pigeonhole principle). By statement $1$, we see that there is a subset whose sum is divisible by $N$. We can then have any $N$ and this proof will hold up.
