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Show that any integer $n>7$ can be written as the sum of $3$s and $5$s exclusively,

i.e., $$ 8= 5+3 \\ 9=3+3+3 \\ 10 = 5+5 \\ 11 = 5+3+3 \\ 12 = 3+3+3+3 $$

So I've started in a couple directions without progress. I think it makes sense to write $$n=3x+5y$$ for $x,y \geq 0$, observing that either $3\mid n$ or $5\mid n$.

It also seems that if you are at $n$, you can increment to $n+1$ by replacing a $5$ with two $3$s. Or you can take $n-1$ and replace a $3$ with a $5$. In this way, it seems like you should be able to keep incrementing after $n=8$, but these ideas aren't formalizing into anything...

Michael Hardy
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Carser
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    This is a special case of this old result : If $m,n$ are co-prime positive integers, each greater than $1,$ then (1) there are no non-negative integers $x,y$ such that $mn-m-n=xm+yn,$ and (2) for every integer $p>mn-m-n $ there exist non-negative integers such that $p=xm+yn.$ This has had so much circulation that it ought to have a name. – DanielWainfleet Aug 31 '16 at 17:17

6 Answers6

10

Hint:

You have an expression for $8, 9$, and $10$.

Now if $n$ can be written as such a sum, what can you say about $n + 3$?

Caleb Stanford
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  • Oh, that makes it seem so simple. So just start with the base case of $8,9,10$, then note that $3|(n+3k)$ for $k\geq0$ starting from the three consecutive integers in the base case? – Carser Aug 31 '16 at 01:46
  • @Carser Well, I don't know about $3 \mid (n + 3k)$. But $3 \mid (3k)$. You have the right idea. – Caleb Stanford Aug 31 '16 at 01:47
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Here is a way to do this using your ideas:

1) As you have shown, 8 can be written in this way.

2) Let $n$ be in $N$ with $n\ge8$ and $n=3x+5y$ where $x,y\ge0$.

a) If $y\ge1$, $\;n+1=3(x+2)+5(y-1)$.

b) If $y=0$, then $3x=n\ge8\implies x\ge3$, so $\;n+1=3(x-3)+5(y+2)$.

Therefore every integer $n\ge8$ can be written in this form by induction.

user84413
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In general, if $a$ and $b$ are two coprime positive integer, every integer $\geq (a-1)(b-1)$ can be written as $ja+kb$ with $j,k\in\mathbb{N}$. This is also known as Frobenius coin (change) problem, see also here.

Community
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Jack D'Aurizio
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You're on the right track.

Set $$k = \left\lfloor \frac{n}{5} \right\rfloor.$$ Then,

  • If $n$ is a multiple of 5, set $x = 0$ and $y = k$ and you're done.
  • If $n \equiv 1 \pmod 5$, set $x = 2$ and $y = k - 1$.
  • If $n \equiv 2 \pmod 5$, set $x = 3$ and $y = k - 2$.
  • If $n \equiv 3 \pmod 5$, set $x = 1$ and $y = k$.
  • If $n \equiv 4 \pmod 5$, set $x = 3$ and $y = k - 2$.

Given the requirement that $x \geq 0$, $y \geq 0$, these five settings work provided $k > 2$.

This is not the most efficient way, but it is sufficient and adequate. You might be able to optimize this by changing the denominator to 3 in the fraction to determine $k$, then coming up with the three settings of $x$ and $y$.

Robert Soupe
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$2*3 - 5 = 1$

$2n*3 - 5n = n$.

$3(2n - 5k) + 5(3k - n)= n$.

To assure that $2n - 5k \ge 0$ and $3k - n > 0$...

If $n = 3m - r; r = 0, 1,2$ then $k$ can be anything equal or greater than $m$ so long as $2n - 5k \ge 0$ i.e. $6m - 2r - 5k \ge 0\implies k \le 6m/5 - 2r/5= m + \frac{m-2r}5$.

So long as $m \ge 4$ we will always be able to find such $k$. i.e. so long as $n \ge 3*4 -2 = 10$.

If $m = 3$ we'll be able find such $k$ if $r \le 1$ i.e. if $n = 8$ or $9$.

We will not be able to find any such $k$ for $n =7$ (where $m = 3; r= 2$).

=======

Or another way:

If we can find $n = 3a + 5b$ we can find $n+1 = 3(a-3) + 5(b+2) = 3(a+2) + 5(b-1)$ so long as either $a \ge 3$ or $b \ge 1$.

If we have $a + b \ge 2$ and $n=3a+5b$ we can find $n+1$ and in doing so either $a$ will increase by $2$ and $b$ (which was at least $1$) decrease by $1$ and $(a+2) + (b-1)\ge 2$ or $a$ (which was at least $3$) will decrease by $3$ but $b$ will in by $2$ so $(a - 3) + (b+2)\ge 2$.

Thus by induction if $n = 3a + 5b; a+b \ge 2$ then $n+1 = 3a' + 5b'; a'+b' \ge 2$.

Base case: $n = 8 = 3*1 + 5*1$. So is possible for all $n \ge 8$.

fleablood
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Since $8=5+3$ we consider those $n$'s for which $n\ge 9$.

  • If $n=3k$ then we are done.

  • If $n=3k+1$ then observe that $3k+1=3(k-3)+2\cdot 5$.

  • If $n=3k+2$ then observe that $n=3k+2=3(k-1)+ 5$.