Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am not understanding how the following formular works

Actually what I did (for my programming homework was loop through the max of 2 numbers until I get 1 that divides both given numbers without a remainder). Bruteforce, so not ideal. So I want to understand more about this formular instead of just using it.

public static int gcd(int a, int b) {
   * Get max of both numbers
   * Loop down max number and test if its divisible by both numbers
   * if yes, thats the answer, else continue
   * if no answer, return 0
  int max = Math.max(a, b);
  for (int i = max; i > 0; i--) {
    if (a % i == 0 && b % i ==0) {
      return i;
  return 0;
share|cite|improve this question
Note that the comment says exactly what the loop does. (a % i takes the remainder after dividing a by i.) – J. M. Aug 23 '11 at 6:11
Note that the Euclidean algorithm article on Wikipedia gives a few better implementations... – J. M. Aug 23 '11 at 6:13
@J.M. I am new to this site and not a maths person so I'm not sure what other tags should I put. I know the formula works. I know % or mod is the remainder. But I do not understand why after recursively dividing the quotient with the remainder (of the prev iteration) will give the gcd of 2 numbers. Hope you get what I mean – Jiew Meng Aug 23 '11 at 6:16
I will take a look at the article soon, gtg to next lesson – Jiew Meng Aug 23 '11 at 6:18
After peering closely at the code... you could have started with the minimum instead of the maximum; $a$ can't be divisible by $b$ is $b$ is bigger, no? The "Description" portion of the wiki-article explains Euclid's idea. – J. M. Aug 23 '11 at 6:25

$a{\rm\ mod\ }b$ is the remainder when you divide $a$ by $b$, so we can say two things about it: it is $a-bq$ for some integer $q$, and it is between $0$ and $b-1$, inclusive.

I want to show $\gcd(a,b)=\gcd(b,a{\rm\ mod\ }b)$.

Let $d=\gcd(a,b)$. Then $d$ divides both $a$ and $b$. So it divides both $a-bq$ and $b$. So it divides $\gcd(b,a{\rm\ mod\ }b)$.

Conversely, let $e=\gcd(b,a{\rm\ mod\ }b)$. Then $e$ divides both $b$ and $a-bq$. So it divides both $a$ and $b$. so it divides $\gcd(a,b)$, and we are done.

Now the only question remaining about the formula is why it always gives an answer, that is, why it can't just keep going forever. Well, supposing $a\gt b$, and comparing $\gcd(a,b)$ and $\gcd(b,a{\rm\ mod\ }b)$, we've decreased both arguments (that is, replacing $a$ with $b$, and replacing $b$ with $a{\rm\ mod\ }b$, we've replaced both with things that are smaller). The way the (positive) integers work, that can't go on forever. It has to stop sometime. But the only way it can stop is when $b=0$ (since if $b\gt0$ you can always keep going), and then $\gcd(a,0)=a$ kicks in.

share|cite|improve this answer

If $\rm\:d\:|\:b\:$ then $\rm\:d\:|\:a\: \iff\: d\ |\ a-n\:b\:.\:$ Thus $\rm\ a,b\ $ and $\rm\ a-n\:b,\:b\ $ have the same common divisors, therefore the same greatest common divisor. Finally $\rm\ a-n\:b\ =\ a\ mod\ b\ $ for some $\rm\:n\in \mathbb Z\:.$

share|cite|improve this answer

The basic idea is this: if you want gcd($a$, $b$) and $a>b$, then we can instead compute gcd($b, a \bmod b$) and we've made progress. It's progress because when $a>b$, we know $a \bmod b$ is smaller than $b$, it's a remainder after all. And if we make positive inputs smaller and smaller, eventually we must terminate.

The real question is this: why is gcd($a$, $b$) the same as gcd($b, a \bmod b$)?

Well, let's answer an easier question. Instead of trying to wrap your head around $a \bmod b$, let's restate the problem. Why is gcd($a$, $b$) the same as gcd($b, a-b$)? This question is almost the same, but we're using minus instead of mod because it's easier to understand when this stuff is new to you. And mod is just a repeated application of minus anyway, right?

So let's prove the "easier" version. Well, if some divisor $d$ goes evenly into $a$ and $b$, then it must go into their difference $a-b$, right? To be more mathematical, if $a = kd$ and $b=jd$, then $a-b = kd-jd = (k-j)d$ and clearly $d$ goes evenly into $(k-j)d$. Also, if $d$ goes evenly into $(a-b)$ then any $d$ dividing $a$ must divide $b$, so we're done.

If this still isn't clear, draw a number line and convince yourself that two multiples of 3 (for example) are always a multiple of 3 apart. Then convince yourself that any multiple of 3 plus a multiple of 3 must also be a multiple of 3.

Then try it for numbers other than 3.

share|cite|improve this answer
One needs both directions: if $d|a$ then $d|b\ \iff\ d|a-b$ – Bill Dubuque Aug 23 '11 at 12:10
@Bill Dubuque: Thanks, Bill. I was trying to stay informal, but I shouldn't sacrifice such an essential part of the justification. Fixed. – Fixee Aug 23 '11 at 17:54

It all rests on one simple fact. If $b = aq + r$ and $b, q, a, r\in \mathbb{Z}$, then $\gcd(a,b) = \gcd(a,r)$. Just show that an integer divides $a$ and $b$ iff it divides $a$ and $r$ and you are done.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.