GCD / LCM

The greatest common divisor of a set of integers is the largest number that divides each integer in the set. We denote the greatest common divisor by gcd(a,b,) \gcd(a, b, \ldots) . We can attempt to find this value by listing all divisors of the integers and finding the largest divisor. However, such a procedure can get tedious.

If we know the prime factorization, then there is a simpler approach. If the prime factorizations of aa and bb are

a=p1α1p2α2pkαk,b=p1β1p2β2pkβk,\begin{aligned} a & = p_1 ^{\alpha_1} p_2 ^{\alpha_2} \ldots p_k ^{\alpha_k}, \\ b & = p_1 ^{\beta_1} p_2 ^ {\beta_2} \ldots p_k ^ {\beta_k}, \\ \end{aligned}

then the GCD of the numbers is equal to

gcd(a,b)=p1min(α1,β1)p2min(α2,β2)pkmin(αk,βk). \gcd(a,b) = p_1 ^{\min(\alpha_1, \beta_1)} p_2 ^{\min(\alpha_2, \beta_2)} \ldots p_k ^{\min(\alpha_k, \beta_k)} .

A similar formula holds for finding the GCD of several integers, by taking the smallest exponent for each prime.

Similarly, the least common multiple of a set of integers is the smallest (positive) number which is a multiple of each integer in the set. We denote this value as lcm(a,b,) \mbox{lcm}(a, b, \ldots). We can attempt to find this value by listing all multiples of the integers in the set, and then finding the one which is the smallest.

However, as above, if we know the prime factorization, then computing the least common multiple is much simpler:

lcm(a,b)=p1max(α1,β1)p2max(α2,β2)pkmax(αk,βk). \mbox{lcm}(a,b) = p_1 ^{\max(\alpha_1, \beta_1)} p_2 ^{\max(\alpha_2, \beta_2)} \ldots p_k ^{\max(\alpha_k, \beta_k)}.

Note: Another approach to find the GCD of 2 numbers is through the Euclidean Algorithm.

Worked Examples

1. Show that the GCD of 2 numbers is indeed equal to

G=p1min(α1,β1)p2min(α2,β2)pnmin(αn,βn). G = p_1 ^{\min(\alpha_1, \beta_1)} p_2 ^{\min(\alpha_2, \beta_2)} \ldots p_n ^{\min(\alpha_n, \beta_n)} .

Solution: Clearly, GG is a divisor of both aa and bb, and so Ggcd(a,b) G \leq \gcd(a,b) .

We will now show that there is no greater common divisor, by considering the prime factorization of gcd(a,b)\gcd(a,b) . Any prime divisor of gcd(a,b)\gcd(a,b) is also a prime divisor of aa and bb. Hence, it is sufficient for us to simply consider the primes that divide aa or bb.

Given any prime pip_i that divides aa or bb, let piαi p_i ^ {\alpha_i} and piβip_i ^{\beta_i} be the largest powers of pp that divide aa and bb respectively. Note that we could have αi,βi=0 \alpha_i, \beta_i = 0. Clearly, pmin(αi,βi)a p^{\min(\alpha_i, \beta_i) } \mid a and pmin(αi,βi)b p^{\min(\alpha_i, \beta_i) } \mid b , hence pmin(αi,βi)gcd(a,b) p^{\min(\alpha_i, \beta_i) } \mid \gcd(a,b) . Furthermore, since pmin(αi,βi)+1 p^{\min(\alpha_i, \beta_i) +1 } does not divide either aa or bb by construction, thus pmin(αi,βi)+1 p^{\min(\alpha_i, \beta_i) +1 } does not divide gcd(a,b) \gcd(a,b) . Hence, gcd(a,b)G \gcd (a,b) \leq G .

As such, we have gcd(a,b)=G \gcd(a,b) = G.

Note: You can use a similar method to prove the claim for lcm(a,b)\mbox{lcm}(a,b) .

 

2. Show that gcd(a,b)×lcm(a,b)=a×b \gcd(a,b) \times \mbox{lcm}(a,b) = a \times b .

First, we show that m+n=min(m,n)+max(m,n) m+ n = \min(m,n) + \max (m,n) . Without loss of generality, nmn\leq m. Hence, min(m,n)=n \min (m,n) = n and max(m,n)=m \max(m,n) = m , thus m+n=min(m,n)+max(m,n) m + n = \min(m,n) + \max (m,n) .

Applying this to each of the pairs m=αi,n=βi m = \alpha_i, n = \beta _i , we get that

a×b=i=1kpiαi+βi=i=1kpimin(αi,βi)+max(αi,βi)=gcd(a,b)×lcm(a,b). a \times b = \prod_{i=1}^k p_i ^{\alpha_i + \beta_i } = \prod_{i=1}^k p_i ^ { \min(\alpha_i, \beta_i) + \max(\alpha_i , \beta_i)} = \gcd(a,b) \times \mbox{lcm}(a,b).

 

3. Given that aa and bb are 2 integers such that 13gcd(a,b)=lcm(a,b) 13 \gcd(a,b) = \mbox{lcm}(a,b) and a+b=2016 a + b = 2016, what are the values of aa and bb?

Let G=gcd(a,b) G = \gcd(a,b) and L=lcm(a,b) L = \mbox{lcm}(a,b).
Let a=aG a = a^* G and b=bG b = b^* G where gcd(a,b)=1 \gcd(a^*, b^*) = 1 by construction. Since (aG)×(bG)=ab=GL=G×13G (a^*G)\times (b^*G) = ab = GL = G \times 13 G , we get that ab=13 a^* b^* = 13 . Hence, we have {a,b}={1,13} \{ a^*, b^* \} = \{ 1, 13\} . WLOG, we may assume that aba\leq b, and thus a=G,b=13G a = G, b = 13 G .

Since 2016=a+b=G+13G=14G 2016 = a + b = G + 13G = 14 G , thus G=201614=144G = \frac{2016} { 14} = 144 . Thus, {a,b}={144,1872} \{a, b\} = \{ 144, 1872 \} .

Note by Arron Kau
5 years, 4 months ago

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From where do you get so much information???

Anuj Shikarkhane - 5 years, 1 month ago

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Books. E.g. Introduction to number theory by CJ Bradley (from UKMT) or Elementary Number Theory by Jones (yellow book).

Curtis Clement - 4 years, 9 months ago

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@Arron Kau What is the least lcm\normalsize lcm of nn numbers adding upto mm. Do we have a subjective approach for it?

Satvik Golechha - 5 years, 1 month ago

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thanks.............

Shatrughna Kumar - 5 years ago

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