How many subset of $A=\{1,2,3...,35\} $ divisible by $5$. How many subset of $A=\{1,2,3...,35\} $ has exeact 26 elements such that sum of all elements divisible by $5$.
I try set: $A(0),A(1),...,A(4)$ are set, which has 26 elements in $A$, satisfy sum of all elements divide $5$ remainder$0,1,2,3,4$.
We have $|A(0)|+|A(1)|+|A(2)|+|A(3)| +|A(4)|=\binom{35}{26}$
That's enought, if we can count each, but I can't to count continous.
 A: Let $\Gamma$ be the set of all $26$-element subsets of $\mathbb{Z}_{35}$. Define the partition
$$ \Gamma=\Gamma_0\sqcup \Gamma_1\sqcup\Gamma_2\sqcup\Gamma_3\sqcup\Gamma_4 $$
where $\Gamma_k\subseteq\Gamma$ consists of those $A\in\Gamma$ with $\sum_{a\in A}a\equiv  k\bmod 5$.
There is a function $\tau$ on $\Gamma$ defined by $\tau(A)=\{a+1\mid a\in A\}$. You can show $\tau(\Gamma_k)=\Gamma_{k+1}$ (with the index interpreted $\bmod 5$). Therefore, each  $\Gamma_k$ has the same size, which must therefore be $1/5$th that of $\Gamma$.
In conclusion, $|\Gamma_0|=\frac{1}{5}\binom{35}{26}$.
A: First we provide a brutal approach with generating functions. This is not meant to be insightful directly (though illustrates generating functions), but with numerics we can make a hint/observation at the very end. Denote
$$
A = \prod_{k=1}^{35} (1+x^kt)= (1+xt)(1+x^2t)(1+x^3t)\cdots(1+x^{35}t)=\sum c_{n,m}x^nt^m.
$$
Here $x^i$ keeps tracks of the use of the integer $i$ in the sum, and the variable $t$ keeps track of how many summands we use. Hence $c_{n,m}$ gives the number of ways to use $m$ many distinct numbers from $\{1,\ldots,35\}$ that sum to $n$. So for our problem seeking sums that are multiples of 5 with 26 summands, we look for the sum of coefficients
$$
\sum _k c_{5k,26}.
$$
Using SageMath(https://sagecell.sagemath.org/) does quick work for us:

var('x','t')
A = 1
for i in [1..35]:
    A = A* (1 + x^i*t)

m = A.expand().coefficient(t^26)
c = 0
for i in [0..150]:
    c = c + m.coefficient(x^(5*i))
c
---
output:
14121492

You can use this value to check your work.
Now to give you a hint/observation:
$$
\frac{35 \choose 26}{14121492} = 5.
$$
So perhaps you can make a guess on $|A(0)|,|A(1)|,|A(2)|,|A(3)|,|A(4)|$.

Hint 2.
We might have guess that $|A(0)| = |A(1)| =|A(2)| = |A(3)| = |A(4)|$, as $14121492$ goes in exactly 5 times into $35 \choose 26$. But how can we prove this? Well if we can establish bijections between these sets $A(i)$, then that shows their sizes are the same. Let us consider an element from $A(0)$, this is a 26-element set from $\{1,\ldots,35\}$ whose sum has remainder 0 modulo 5. Say they are
$$
a_1 , a_2 , \ldots, a_{26}.
$$
Now consider a new 26-element set where we add 1 to each of the elements above:
$$
a_1 +1 , a_2 +1 , \ldots, a_{26}+1
$$
What can we say about this new set? If we add these numbers up, we get a sum of $26 \equiv 1\mod 5$, which means this is an element from $A(1)$!. (Small detail, if $a_i+1$ exceeds 35, then we consider the number mod 35). This process is reversible by subtracting 1 from each element in an element of $A(1)$. Thus $|A(0)|=|A(1)|$. The rest can be shown similarly to show $|A(0)| = |A(i)|$ by adding $i$ to each entry like before.
Remark. This worked out nicely like this because we happen to have $gcd(26,5)=1$. If instead you are looking for sums that are divisible by 5 with using exactly 25 elements instead, then we won't have such nice simple bijections.
