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 $\gcd(a, b, \ldots)$. 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 $a$ and $b$ are

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

then the GCD of the numbers is equal to

$\gcd(a,b) = p_1 ^{\min(\alpha_1, \beta_1)} p_2 ^{\min(\alpha_2, \beta_2)} \ldots p_k ^{\min(\alpha_k, \beta_k)} .$

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 $\mbox{lcm}(a, b, \ldots)$. 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:

$\mbox{lcm}(a,b) = p_1 ^{\max(\alpha_1, \beta_1)} p_2 ^{\max(\alpha_2, \beta_2)} \ldots p_k ^{\max(\alpha_k, \beta_k)}.$

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

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

$G = p_1 ^{\min(\alpha_1, \beta_1)} p_2 ^{\min(\alpha_2, \beta_2)} \ldots p_n ^{\min(\alpha_n, \beta_n)} .$

Solution: Clearly, $G$ is a divisor of both $a$ and $b$, and so $G \leq \gcd(a,b)$.

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

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

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

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

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

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

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

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

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

Let $G = \gcd(a,b)$ and $L = \mbox{lcm}(a,b)$.

Let $a = a^* G$ and $b = b^* G$ where $\gcd(a^*, b^*) = 1$ by construction. Since $(a^*G)\times (b^*G) = ab = GL = G \times 13 G$, we get that $a^* b^* = 13$. Hence, we have $\{ a^*, b^* \} = \{ 1, 13\}$. WLOG, we may assume that $a\leq b$, and thus $a = G, b = 13 G$.Since $2016 = a + b = G + 13G = 14 G$, thus $G = \frac{2016} { 14} = 144$. Thus, $\{a, b\} = \{ 144, 1872 \}$.

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## Comments

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TopNewestFrom where do you get so much information???

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

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@Arron Kau What is the

least$\normalsize lcm$ of $n$ numbers adding upto $m$. Do we have a subjective approach for it?Log in to reply

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

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