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# Greatest common divisor/Tutorials

### From Citizendium, the Citizens' Compendium

## Explicit example

For example, the divisors of 60 are

- 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60,

and the divisors of 72 are

- 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72.

The **common divisors** of 60 and 72 are the numbers that appear in both lists:

- 1, 2, 3, 4, 6, 12.

The **greatest common divisor** of 60 and 72 is therefore 12. One writes "gcd(60, 72) = 12", or simply "(60, 72) = 12".

The greatest common divisor is used in reducing fractions to lowest terms, thus:

## An algorithm involving prime factorizations

One method of finding the greatest common divisor of two integers involves factoring both into prime numbers:

Observe that the prime number 2 occurs twice in the factorization of 60, and three times in the factorization of 72. One says that the *multiplicity* of that prime factor is 2 in the first case and 3 in the second. ("Multiplicity" means the number of times the prime factor appears in the factorization.) For each prime factor, one sees which multiplicity is *smaller*: the one in the factorization of 60 or the one in the factorization of 72:

- In the factorization of 60, the multiplicity of the prime factor 2 is 2.
- In the factorization of 72, the multiplicity of the prime factor 2 is 3.
- The smaller of those multiplicities is 2.

Next:

- In the factorization of 60, the multiplicity of the prime factor 3 is 1.
- In the factorization of 72, the multiplicity of the prime factor 3 is 2.
- The smaller of those multiplicities is 1.

Next:

- In the factorization of 60, the multiplicity of the prime factor 5 is 1.
- In the factorization of 72, the multiplicity of the prime factor 5 is 0.
- The smaller of those multiplicities is 0.

For completeness, note that

- all other prime numbers have multiplicity 0 in the factorization of both 60 and 72.
- The smaller of those two multiplicities (both 0) is 0.

Therefore

- the multiplicity of 2 in the factorization of the gcd is 2;
- the multiplicity of 3 in the factorization of the gcd is 1;
- the multiplicity of any other prime number in the factorization of the gcd is 0.

The gcd is therefore 2 × 2 × 3 = 12.

Here is another example: the problem is to find gcd(14280, 1638). We have

- 14280 = 2 × 2 × 2 × 3 × 5 × 7 × 17.
- 1638 = 2 × 3 × 3 × 7 × 13.

For the prime number 2, the multiplicities are 3 and 1; the smaller of those is 1.

For the prime number 3, the multiplicities are 1 and 2; the smaller of those is 1.

For the prime number 5, the multiplicities are 1 and 0; the smaller of those is 0.

For the prime number 7, the multiplicities are 1 and 1; the smaller of those is 1.

For the prime number 11, the multiplicities are 0 and 0; the smaller of those is 0.

For the prime number 13, the multiplicities are 0 and 1; the smaller of those is 0.

And so on—for all larger prime numbers the multiplicity is 0.

Therefore gcd(14280, 1638) = 2 × 3 × 7 = 42.

## An algorithm not involving prime numbers

When the prime factors of a number are large, the algorithm above may be inefficient. Euclid's algorithm does not involve prime factorizations and runs fast in such cases. It may be described as follows:

- (1) Replace the larger of the two numbers with the remainder on division of the larger by the smaller.
- (2) Repeat step (1) until one of the two numbers is 0.
- (3) The one that is not 0 is the gcd.

For an example, let's find the greatest common divisor of the last pair of numbers we used above, (14280, 1638).

- 14280 / 1638 = 8 remainder 1176. because 1638 times 8 is 13104.

- Now we take the new pair of numbers (1638, 1176).
- 1638 / 1176 = 1 remainder 462

- Now we have (1176,462).
- 1176 / 462 = 2 remainder 252

- Now we have (462, 252).
- 462 / 252 = 1 remainder 210.

- Now we have (252,210).
- 252 / 210 = 1 remainder 42.

- Now we have (210,42).
- 210 / 42 = 5 remainder 0.

- Now we have (42,0).

Therefore gcd(14280,1638) = 42.