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It is well-known that given two primes $p$ and $q$, $pZ + qZ = Z$ where $Z$ stands for all integers. It seems to me that the set of natural number multiples, i.e. $pN + qN$ also span all natural numbers that are large enough. That is, there exists some $K>0$, such that $$pN + qN = [K,K+1,...).$$

My question is, given $p$ and $q$, can we get a upper bound on $K$?

Bill Dubuque
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3 Answers3

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$K = pq + 1$ if $\mathbb{N} = \{ 1, 2, 3, ... \}$, and $K = pq - p - q + 1$ if $\mathbb{N} = \{ 0, 1, 2, 3, ... \}$. This is known as the coin problem, or Frobenius problem (and you only need $p, q$ relatively prime). It frequently appears on middle- and high-school math competitions.

Edit: I completely misremembered how hard the proof is. Here it is. If $n$ is at least $pq+1$, then the positive integers

$$n-p, n-2p, ... n-qp$$

have distinct residue classes $\bmod q$, so one of them must be divisible by $q$. On the other hand, it's not hard to see that $pq$ itself cannot be written in the desired way.

Qiaochu Yuan
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  • Qiaochu: Typo? $K=pq-p-q+1$. –  Oct 28 '10 at 15:16
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    @Byron: this would be true if the OP is allowing non-negative integer multiples, but the OP wants positive integer multiples. – Qiaochu Yuan Oct 28 '10 at 15:19
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    @Qiaochu: the OP said "nautral numbers". Please tell me you're not one of those misguided souls who doesn't think $0$ is a natural number! (More seriously, you should probably clarify this in your response. You might as well give the answer both ways.) – Pete L. Clark Oct 28 '10 at 16:22
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    @Pete: I'm agnostic. Generally I use Z_{\ge 0} for the non-negative integers. I'll add the clarification. – Qiaochu Yuan Oct 28 '10 at 16:29
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    Nice answer. However, to ensure that the coefficient of p is positive, I think you should be saying that n-p,n-2p,...,n-qp have distinct residue classes mod q. – George Lowther Nov 06 '10 at 20:59
  • The proof for the case $K = pq - p - q + 1$ is given here: https://math.stackexchange.com/questions/66963/largest-integer-that-cant-be-represented-as-a-non-negative-linear-combination-o?noredirect=1&lq=1 – 19021605 Aug 08 '25 at 15:18
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Here is a $3$-line arithmetical proof (with geometric view). Below is an inductive variant (highly recommended for motivation / warm-up: first browse cases $p,q\!=\!3,5$ and $p,q\!=\!17,23$ below).

Example $ $ Let $\,f(x,y)=17x+23y.\,$ The gcd $\rm\color{#0a0}{Bezout}$ identity here is $\,\color{#0a0}{f(-4,3)=1}\,$ so if $\,f(x,y) = n\,$ then we can add $\:\!1,\:\!$ via $\,f(\color{#c00}{x\!-\!4},y\!+\!3) = f(x,y)+\color{#0a0}{f(-4,3)} = n+\color{#0a0}1,\,$ and if $\,\color{#c00}{x-4\le 0}\,$ then we can make it $\:\!> 0\:\!$ by further $\rm\color{#c00}{adding\ (23,-17)},\,$ by $\:\!f(23,-17)=0,\,$ e.g.

$391 = f(\ \ \ 0,17)\ $
$392 = f(\color{#c00}{-4},20)\ \ \ $ by adding $\ 1 = f(-4,3)\,$
$\qquad\! = f(\ 19,\ \ 3)\ \ \ $ by $\color{#c00}{{\rm adding}\:\!\ 0 = f(23,-17)}\ $ to make $\,x>0$
$393 = f(\ 15,\ \ 6)\ \ \ $ by adding $\ 1 = f(-4,3)$
$394 = f(\ 11,\ \ 9)\qquad\quad\cdot $
$395 = f(\ \ \ 7,12)\qquad\quad\cdot$
$396 = f(\ \ \ 3,15)\qquad\quad\cdot$
$397 = f(\color{#c00}{-1},18)\qquad\quad\cdot$
$\qquad\! = f(\ \,22,\ 1)\ \ \ $ by $\color{#c00}{{\rm adding}\:\!\ 0 = f(23,-17)}\ $ to make $\,x>0$

Below we give an inductive proof showing that the same method works generally.

Let $f(x,y) = px+qy.\,$ $f(a,b)=pa+qb=1\,$ for some $\,a,b\in\Bbb Z\,$ by Bezout & $\,p,q\,$ coprime. Wlog we may assume that $\,\color{#c00}{-q < a}< 0\, $ (so $\,\color{#0a0}{b>0})\,$ by choosing $\,a = (p^{-1}\!\bmod q) \color{#c00}{- q}$.

From base case $\,f(0,p) = pq\,$ we inductively apply: $ $ if $\,f(x,y) = n \ge pq\,$ with $\,x,y\ge0\,$ then there are $\,\bar x,\bar y >0\,$ with $\,f(\bar x,\bar y) = n\!+\!1,\,$ by adding $\,(a,b)\,$ to $\,(x,y)\,$ to increment $\,n,$ then if $\,a\!+\!x \le 0,\,$ add $\,(q,-p)\,$ to force $\,\bar x,\bar y>0.\,$ We prove it works, using $f\,$ is linear & increasing.

$x\! +\! a > 0\Rightarrow f(\bar x,\bar y)\!=\! f(x\!+\!a,y\!+\!b) = f(x,y)\!+\!f(a,b) = n\!+\!1;\,$ $\, y\ge0,\color{#0a0}{b>0}\Rightarrow y\!+\!b>0$

$y\!+\!b > p\Rightarrow f(\bar x,\bar y) \!=\!f(x\!+\!a\!+\!q,y\!+\!b\!-\!p)=f(x,y)\!+\!f(a,b)\!+\!f(q,-p) = n\!+\!1\!+\!0,\,$ and $\,x\ge0,\,\color{#c00}{a\!+\!q}>0\Rightarrow x\!+\!a\!+\!q>0.\,$ This case must hold if the prior fails (else both fail so

$\begin{align}&x\!+\!a\le 0\\ &y\!+\!b\,\le p\end{align}\Rightarrow\, f(x\!+\!a,y\!+\!b)\le f(0,p)\,$ i.e $\:n\!+\!1 \le pq,\,$ contra $\,n \ge pq\,$ by hypothesis).

Remark $ $ A unit shift transforms to the case of nonnegative (vs. above) positive solutions $\,x,y$.

There is much literature on this classical problem. To locate such work one should search on the many aliases, e.g. postage stamp problem, Sylvester/Frobenius coin problem, Diophantine problem of Frobenius, Frobenius conductor, money changing, coin changing, change making problems, chicken McNugget theorem, h-basis and asymptotic bases in additive number theory, integer programming algorithms and Gomory cuts, knapsack problems and greedy algorithms, etc.

Bill Dubuque
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This follows easily by Bézout's lemma (as Qiaochu notes, x and y only need to be coprime to generate the unit ideal (indeed, this is the proper generalization of the term "coprime")).

We can ever so-slightly strengthen this to show that $\mathbf{Z}$ is a PID (this is the true content of Bézout's lemma).