Prove that all the natural numbers except 1 can be expressed in the form of 2x + 3y

Question

Is this statement true? Statement: "All the natural numbers except 1 can be expressed in the form of $2x + 3y$"

NOTE: $x$ and $y$ should be non-negative

If yes then can you show how to prove this?

Context: There is a programming question, solution of which can be optimized using this property.

Problem statement:

You are given a $0$-indexed integer array tasks, where tasks[i] represents the difficulty level of a task. In each round, you can complete either $2$ or $3$ tasks of the same difficulty level. Return the minimum rounds required to complete all the tasks, or $-1$ if it is not possible to complete all the tasks.

Solution which uses this property.

Answer 1

Let $n$ be a natural number and $n>1$. There are two possibilitites, $n$ is even or $n$ is odd. If $n$ is even then $n=2x$ for some integer $x$. If $n$ is odd then $n\geq 3$, and so $(n-3)$ is a non-negative integer which is even, therefore $(n-3) = 2x$ and so $n = 2x + 3$. Regardless of what $n$ is we can see we can always express $n$ in the form $2x+3y$ where $x\geq 0$ and $y\geq 0$ (in fact, with the stronger requirement that $y=0$ or $y=1$).

Answer 2

Assuming we take $x$ and $y$ to be non-negative integers, we can proceed quite directly.

Let $n$ represent an arbitrary natural number we'd like to represent in this fashion. The case where $n$ is even works out rather immediately, since if $n$ is even then $n = 2k$ for some positive integer $k,$ so let $x = k, y = 0$ and $2(k) + 3(0) = 2k = n.$

If $n$ is odd, then $n = 2k + 1$ for some integer $k,$ and $n \geq 3$ implies $k \geq 1.$ We know the odd $1$ left over has to come from a $3,$ so we can rearrange our form for $n$ into $n = 2k + (- 2 + 2) + 1 = (2k - 2) + 3 = 2(k-1) + 3(1),$ and because $k \geq 1$ we have $k - 1 \geq 0,$ so $x = k - 1, y = 1$ gives us nonnegative solutions for $x$ and $y.$

Now because all natural numbers are either even or odd, and the only restriction on our natural value $n$ was that it was greater than $1,$ this concludes the proof that all natural numbers greater than $1$ can be represented in this form.

However, this does not guarantee that the breakdowns given in this proof are optimal. For instance, these procedures would break $n = 6$ into three twos, when it's rather clear the optimal grouping would be two threes. In general, we should be able to see that the smallest combined value for $x$ and $y$ comes from when we maximize the number of $3$s used, instead of $2$s as we are in our procedure. We can adjust for this by simply shifting from letting the number of $3$s determine the remainder mod $2,$ we can let the $2$s determine the remainder mod $3,$ and fill the rest with as many $3$s as possible.

In case this is a homework problem I'll hold off from undergoing this process myself for now, but I would suggest carrying out the analogue of the odd-even argument with the remainders of $0, 1,$ and $2$ mod $3$ instead. This will give you representations which maximize $y,$ which I claim would minimize $x + y,$ the total number of rounds. (leaving out the proof for now for the sake of time)

Answer 3

Thanks everyone for the amazing proofs!

For the optimization part we have to see number in the form of mod $3$ and not in mod $2$.

We can see that to optimize our solution we need to use $3$ as many times as possible and the remaning part will be made up using the $2$.

So instead of bifurcating the number as even and odd which is in the form of mod $2$ ($2k$ or $2k + 1$) we will represent our number in the form of $3K$ or $3K + 1$ or $3K + 2$.

Cases:

• $n = 3K$ we need only $K$ operations and this will by default be the minimum number of operations required
• $n = 3K + 1$ we will reframe this as $3(k - 1) + 3 + 1 = 3(K - 1) + 2 + 2$ which means we require $K + 1$ operations ($K - 1$ operations for using $3$ and $2$ operations for using $2$).
• $n = 3K + 2$ this also means we need $K + 1$ operations ($K$ times for $3$ and $1$ time for using $2$).