How to sort numbers from 1 to 20 into 6 piles where sum in each pile is the same? 
How to sort numbers from 1 to 20 into 6 piles where sum in each pile is the same?

This question, my son got in school and I can't figure out what is the correct approach to solve this.
 A: If $n$ is the sum of one pile, then we must have
$$1+2+3+\ldots+19+20=6n\;.$$
The sum on the left is $\dfrac{20\cdot21}2=210$, so we must have $6n=210$ and $n=35$. Now you just have to break the integers from $1$ through $20$ into groups that total $35$ each. You can do this in any way you like. For instance, we could start by letting one group be $\{15,20\}$. Another could then be $\{16,19\}$, and $\{17,18\}$ could be a third. That leaves the integers from $1$ through $14$ to be divided into three groups. $14+13=27$, and $35-27=8$, so you could let the fourth group be $\{8,13,14\}$, leaving $\{1,2,3,4,5,6,7,9,10,11,12\}$ to be split into two groups.
But it really doesn’t matter how you do it: just keep forming groups that sum to $35$, and when you’ve formed five of them, whatever numbers remain must sum to $35$ to give you your sixth group.
A: The sum of the numbers from $1$ to $20$ is
$\left(\begin{array}{c}
20 + 1 \\
+ \\
19 + 2 \\
+ \\
18 + 3 \\
+ \\
17 + 4 \\
+ \\
16 + 5 \\
+ \\
15 + 6 \\
+ \\
14 + 7 \\
+ \\
13 + 8 \\
+ \\
12 + 9 \\
+ \\
11 + 10 \\
\end{array}\right)
=
\left(\begin{array}{c}
21 \\
+ \\
21  \\
+ \\
21  \\
+ \\
21  \\
+ \\
21  \\
+ \\
21  \\
+ \\
21 \\
+ \\
21  \\
+ \\
21 \\
+ \\
21 \\
\end{array}\right)
=10\times 21 = 210$
So each pile must sum to $210 \div 6 = 35$
One thing that might make this problem look more difficult than it is, is that you might think that there is only one solution. There are lots of solutions. So don't worry too much about what effect the numbers you choose now will affect the choices you will have to make later.
So let's start by listing all of the pairs of numbers that add up to $35$


*

*$(20, 15)$

*$(19, 16)$

*$(18, 17)$


That's all of the pairs of numbers that add up to $35$
That leaves $1,2,3,4,5,6,7,8,9,10,11,12,13,14$ which need to be split up into three groups that add up to $35$.
So let's start by finding three of the remaining numbers that add up to $35$.


*

*$(14, 13, 8)$


That leaves $1,2,3,4,5,6,7,9,10,11,12$ which need to be split up into two more groups that add up to $35$.
So let's look for three more of the remaining numbers that add up to $35$.
Darn, there aren't any. So let's look for four of the remaining numbers that add up to $35$.


*

*$(12, 11, 10, 2)$


will add up to $35$
So we now have five groups of numbers that add up to $35$ and we have these numbers left


*

*$(1,3,4,5,6,7,9)$


Guess what. The remaining seven number HAVE TO add up to $35$. So we are done. the complete list is


*

*$(20, 15)$

*$(19, 16)$

*$(18, 17)$

*$(14, 13, 8)$

*$(12, 11, 10, 2)$

*$(1,3,4,5,6,7,9)$

