20 Gifts and four stacks, How many Gifts in each pile? There are twenty gifts stacked up into $4$ piles.
The first pile has $3$ less than the second pile.
The second pile has $2$ more than the third pile.
The fourth pile has twice as many as the second pile.
How many gifts are in each pile?
This is for my $8$ year old son... Embarassing that I can't figure it out.
 A: At that age they are pretty clever, but usually innocent of algebra. So the expected approach is trial and error, in the schools often called "guess and check." But one might as well be reasonably smart about it. 
Let's visualize the piles of gifts. The second pile is bigger than its two neighbours, and the fourth is twice as big as the second. So the fourth is quite big, and has an even number of gifts. Let's guess that the fourth has $12$ gifts. Then the second has $6$, the third has $4$, and the first has $3$. Add up, we get $25$. Too bad, not right. 
So $12$ for the fourth is too big. Let's try for $10$. Then the second has $5$, the third has $3$, the first has $2$. Add up. Bingo! 
One can work in this way from any of the piles. For example, the first is clearly the smallest. If we guess it is of size $1$, we get $1, 4, 2, 8$, too small a sum. But $2, 5, 3, 10$ get us to the right place.   
A: Let
$$\begin{align*}
A & =\text{number of gifts in the first pile}\\
B & =\text{number of gifts in the second pile}\\
C & =\text{number of gifts in the third pile}\\
D & =\text{number of gifts in the fourth pile}
\end{align*}$$
Then the information given is that
$$A=B-3,\quad B=C+2,\quad D=2B,\quad A+B+C+D=20.$$
We can re-express $B=C+2$ as $C=B-2$. Replacing term by term:
$$\underbrace{A}_{B-3}+B+\underbrace{C}_{B-2}+\underbrace{D}_{2B}=20$$
we see that
$$(B-3)+B+(B-2)+2B=5B-5=20$$
which implies $B=5$, and then $A=2$, $C=3$, and $D=10$.
A: Besides André’s approach, I can also imagine a bright child starting with an empty first pile, three gifts in the second pile, and one gift in the third, easily fulfilling the first three requirements, and then working out that there must be six gifts in the fourth pile. That’s the smallest arrangement that meets the requirements on the relative sizes of the piles; unfortunately, it uses only $10$ of the $20$ gifts.
The first three piles change in step with one another: add one gift to one of them, and you clearly must add one gift to each of them. And every time you add one gift to the second pile, you must add two gifts to the fourth pile. Thus, adding one gift to any of the first three piles forces you to add a total of five gifts. Aha! Do this twice, and you’ve used all $20$ gifts. You’ve now added two gifts to each of the first three piles and four gifts to the fourth pile, so the piles now contain $2$, $5$, $3$, and $10$ gifts, respectively.
