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 $2$.