Greatest common divisor: Difference between revisions
imported>Catherine Woodgold (→An algorithm not involving prime numbers: Adding an example of Euclid's algorithm. Arithmetic either here or in previous section must be wrong.) |
imported>Catherine Woodgold (→An algorithm involving prime factorizations: Correcting a mistake in the gcd calculation) |
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For the prime number 5, the multiplicities are 1 and 0; the smaller of those is 0. | 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. | And so on—for all larger prime numbers the multiplicity is 0. | ||
Therefore gcd(14280, 1638) = 2 × 3 = | Therefore gcd(14280, 1638) = 2 × 3 × 7 = 42. | ||
== An algorithm not involving prime numbers == | == An algorithm not involving prime numbers == |
Revision as of 13:36, 12 May 2007
In arithmetic, the greatest common divisor (gcd) of several integers is the largest number that is a divisor of all of them.
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.
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. 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.
- 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.
Example:
- (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.