TITLE: Is there another simpler method to solve this elementary school math problem?
QUESTION [156 upvotes]: I am teaching an elementary student. He has a homework as follows.

There are $16$ students who use either bicycles or tricycles. The total
  number of wheels is $38$. Find the number of students using bicycles.

I have $3$ solutions as follows.
Using a single variable.
Let $x$ be the number of students in question. The number of students using tricycles is $16-x$. The total number of wheels is the sum of the total number of bicycles times $2$ and the total number of tricycles times $3$.
$$
2\times x + 3 \times (16-x) = 38
$$
The solution is $x=10$.
Using $2$ variables.
Let $x$ and $y$ be the number of students using bicycles and tricycles, respectively.
It implies that
\begin{align}
x+y&=16\\
2x+3y&=38
\end{align}
The solution is $x=10$ and $y=6$.
Using multiples
The multiples of $2$ are $2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22,24, 26,28,30,32,\dotsc$
The multiples of $3$ are $3,6,9,12,15,18,21,24,27,30,33,36,\dotsc$ 
The possible wheel combinations with format $(\#\text{bicycle wheels}, \#\text{tricycle wheels})$:
$\hspace{6cm} (32,6)$ but there are $18$ students
$\hspace{6cm} (26,12)$  but there are $17$ students
$\hspace{6cm} (20,18)$ there are $16$ students
$\hspace{6cm} (14, 24)$ there are $15$ students
$\hspace{6cm} (8, 30)$ there are $14$ students
$\hspace{6cm} (2,36)$ there are $13$ students
Thus the correct combination is $10$ bicycles and $6$ tricycles.
My question
Is there any other simpler method?

REPLY [4 votes]: If $x$ students have bicycles, the total number of wheels is
$$
2x+3(16-x).
$$
To find $x$, we can give an initial guess, like $x=0$, and then compute how far we are from the solution. The squared distance is
$$
(38-2x-3(16-x))^2 = (x-10)^2 \ge 0.
$$
We can continuously update the estimate of $x$ in order to minimize the squared distance by using a differential equation
$$
\frac{dx}{dt} = -2(x-10). 
$$
The solution will converge to the stable solution given by 
$$
\frac{dx}{dt} = 0.
$$
Solving $-2(x-10)=0$, the solution is found to be $x=10$.
(the author asked for a differential solution in the comment...)