Finding all unordered pairs $a,b$, knowing $\gcd(a,b)=12$ and $\operatorname{lcm}(a,b)=360$

Question

If we know that $\gcd(a,b)=12$ and $\operatorname{lcm}(a,b)=360$, how can we find all unordered pairs $a,b$?

The answer in the back of the book I got this from gives $(12,360), (24,180), (36,120), (60,72)$ as the solutions.

Answer 1

When solving such questions, it is best to think in terms of prime factorizations instead of integers. So we start by prime-factorizing $360$ and $12$: \begin{align*} 360 &= 2^3 \cdot 3^2 \cdot 5^1 \\ 12 &= 2^2 \cdot 3^1 \end{align*} $a$ and $b$ must divide $360$, since $360$ is the LCM, so they cannot have any prime factors other than $2, 3, 5$. So \begin{align*} a &= 2^x 3^y 5^z \\ b &= 2^{x'} 3^{y'} 5^{z'} \end{align*} Now from LCM and GCD being $360$ and $12$, we get \begin{align*} \text{max}(x,x') = 3, \text{min}(x,x') = 2 \implies \{x,x'\} = \{2,3\} \\ \text{max}(y,y') = 2, \text{min}(x,x') = 1 \implies \{y,y'\} = \{1,2\} \\ \text{max}(z,z') = 1, \text{min}(z,z') = 0 \implies \{z,z'\} = \{0,1\} \\ \end{align*} So to write out all possibilities, we just have to assign $2$ and $3$ to $x$ and $x'$, $1$ and $2$ to $y$ and $y'$, and $0$ and $1$ to $z$ and $z'$. Since we only want unordered possibilities, we can start out taking $z = 0$, $z' = 1$. So all unordered possibilities for $(a,b)$ are, as you wrote: \begin{align*} (2^2 3^1, 2^3 3^2 5^1) &= (12, 360) \\ (2^3 3^1, 2^2 3^2 5^1) &= (24, 180) \\ (2^2 3^2, 2^3 3^1 5^1) &= (36, 120) \\ (2^3 3^2, 2^2 3^1 5^1) &= (72, 60). \\ \end{align*}

Answer 2

Use the fact that the product of two numbers $a$ and $b$ is the same as the product of gcd$(a,b)$ and lcm$(a,b)$. So looking at alternative factorizations should lead to the answer.

Here we are given $12$ and $360$. Divide $360$ by some factor $k$ and multiply $12$ by that same $k$. That is $a=360/k, b=12k$ for various divisors $k$ of 360.

Answer 3

The problem is equivalent to finding all pairs $(a,b)$ with $gcd(a,b)=1$ and $lcm(a,b)=30$. Indeed, given such a pair, $(12a,12b)$ satisfies your requirements and conversely if you have a pair satisfying your conditions then $(\frac{a}{12},\frac{b}{12})$ satisfies the present condition. Write $30$ as $2\times 3\times 5$. Now the tuples $(a,b)$ satisfying the condition are in bijection with the partitions of $\{2,3,5\}$ into two parts. This is because $a,b$ cannot share a common factor, and their least common multiple has all the factors of $30$. So you decide which are the prime factors of $a$, and then $b$ has to take all the rest.

For example, take $a=2\times 5$, $b=3$, this gives the pair $(10,3)$, and to get one of the tuples in your question we multiply by $12$ and get $(120,36)$.

Answer 4

These problems are sometimes solved with Venn diagrams in middle school, and I think the method is helpful to work through and understand! For an example discussion, see:

Feldman, Z. (2014). Rethinking factors. Mathematics Teaching in the Middle School, 20(4), 230-236.

You can draw a Venn diagram with each circle representing one of $a$ and $b$, and place their shared prime factors in the overlapping region to correspond to their GCF. Once the remaining factors are placed in the other regions, you can multiply across all visible prime factors to get their LCM.

For this problem:

The question now becomes, how can you place the remaining factors in $a$ and $b$ so that the product across all present numbers is $360$?

Already present is a $12$, so the remaining product needs to be $30$ since $12 \cdot 30 = 360$.

Now consider all factor pairs for $30$:

$$(1, 30); (2, 15); (3, 10); (5, 6); (6, 5); (10; 3); (15, 2); (30, 1)$$

For each such factor pair $(m, n)$ you can put $m$'s factors into $a$'s region, and $n$'s factors into $b$'s region. This gives a total of eight solutions (or four solutions, up to symmetry, which are exactly the ones that you listed in your question). As an example, if you take the second factor pair, $(2, 15)$, you could put $2$ into $a$'s region, and $15 = 3 \cdot 5$ into $b$'s region; this results in the following picture:

Computing each region, we find $a = 2 \cdot 2 \cdot 2 \cdot 3 = 24$ and $b = 3 \cdot 5 \cdot 2 \cdot 2 \cdot 3 = 180$.

So in this case, we find that $(a, b) = (24, 180)$.

Going through the remaining cases for each of $30$'s factor pairs yields the other solutions.

Answer 5

If $\gcd(a,b) = 12$, then $12 \mid a$ and $12 \mid b$. Hence $\dfrac{a}{12}$ and $\dfrac{b}{12}$ are integers and $\gcd\left( \dfrac{a}{12}, \dfrac{b}{12} \right) = 1$ . Hence...

\begin{align} ab &= \gcd(a,b) \times \operatorname{lcm}(a,b) \\ ab &= 12 \times 360 \\ \dfrac{a}{12} \dfrac{b}{12} &= \dfrac{360}{12} \\ \dfrac{a}{12} \dfrac{b}{12} &= 30 \\ \end{align}

So, if you are looking for unordered pairs, $\left(\dfrac{a}{12}, \dfrac{b}{12} \right) \in \{ (1,30), (2,15), (3,10), (5,6)\}$

Note that the condition $\gcd\left( \dfrac{a}{12}, \dfrac{b}{12} \right) = 1$ has been satisfied in each case.

So, $(a,b) \in \{ (12,360), (24,180), (36,120), (60, 72)\}$

Answer 6

I think you know {L.C.M }x {H.C.F.}= product of numbers [valid for 2 numbers only].

In your case {L.C.M }x {H.C.F.} =360*12 = a*b N

Note that none of a and b can be smaller than 12 since it is {H.C.F.}. Now only thing you have to do is to resolve 360*12 into two factors. You will find that these ways are (12*360), (24*180), (36*120) and (60*72).

The others will be excluded(such as 48*90) because one of them is not divisible by 12 which is {H.C.F.}.

Hope you get my message, if any problem regarding solution leave comment

Answer 7

Quick Solution

This problem ties in nicely with prime factorisation and multisets. We can count the number of unordered pairs using Pihedron's Theorem and generate them in a similar way to how Caleb used min and max. However, I believe my method is faster. It relies on representing a natural number as a multiset of its prime factors. For example, $12$ can be represented as $2^2 \times 3^1$.

1. Divide LCM by GCD to get the multiset symmetric difference as $\frac{360}{12}=30$.
2. Convert $30$ to a product of its prime factors $2^1 \times 3^1 \times 5^1$.
3. Since $30$ has $3$ distinct prime factors, we know there must be $2^{3} = 8$ ordered pairs.
4. Therefore, there will be 4 unordered pairs.
5. We can generate all intermediate unordered pairs by raising each prime to $0$ or its highest power (which is $1$ for all of the prime factors of $30$ in this case).
   • $(2^0 \times 3^0 \times 5^0, 2^1 \times 3^1 \times 5^1) = (1, 30)$
   • $(2^1 \times 3^0 \times 5^0, 2^0 \times 3^1 \times 5^1) = (2, 15)$
   • $(2^1 \times 3^1 \times 5^0, 2^0 \times 3^0 \times 5^1) = (6, 5)$
   • $(2^1 \times 3^0 \times 5^1, 2^0 \times 3^1 \times 5^0) = (10, 3)$
6. We multiply the pairs by $\gcd(a, b)$ which is $12$.
   • $(1,30) \times 12 = (12, 360)$
   • $(2,15) \times 12 = (24, 180)$
   • $(6,5) \times 12 = (72, 60)$
   • $(10,3) \times 12 = (120, 36)$

Proof

Pihedron's Theorem

We begin by asserting:

• A natural number $x$ can be represented as a multiset of primes.
• The product of all the primes raised to their multiplicity in the multiset equals $x$.

Let $A, B$ be the multiset representation of $a, b$ respectively and $f$ be a function that maps a multiset of primes to its respective natural number. Therefore:

• $\text{lcm}(a,b) = A \cup B$
• $\gcd(a,b) = A \cap B$
• $f(A \operatorname\Delta B) = \frac{\text{lcm}(a,b)}{\gcd(a,b)}$ because $(A \cap B) \cup (A \operatorname\Delta B) = A \cup B$

For each prime in $(A \operatorname\Delta B)$, we must assign it to either $\Delta A$ or $\Delta B$. We must not assign the same prime to both multisets even if it has multiplicity greater than $1$ because it will overestimate the size of $A \cap B$ by introducing a new common factor. In other words, $\gcd(f(\Delta A), f(\Delta B)) = 1$ must be true. The number of ways we can split $A \operatorname\Delta B$ into an ordered multiset pair can be computed using the combinations with replacement formula $2^{n(A \operatorname\Delta B)}$ where $n(X)$ counts the number of distinct elements in the multiset $X$. The number of distinct prime factors of $x$ is denoted by $\omega(x)$. Therefore:

• $n(A \operatorname\Delta B) = \omega\left(\frac{\text{lcm}(a,b)}{\gcd(a,b)}\right)$
• There are $2^{n(A \operatorname\Delta B)}$ or $2^{\omega\left(\frac{\text{lcm}(a,b)}{\gcd(a,b)}\right)}$ ordered pairs of $(\Delta A, \Delta B)$ that satisfy the co-prime condition.
• There are half as many unordered pairs except when $\text{lcm}(a,b) = \gcd(a,b)$ due to $A \operatorname\Delta B$ only being able to form one symmetric pair.
• All ordered pairs satisfying $\text{lcm}(a,b)$ and $\gcd(a,b)$ can be generated by multiplying every $(f(\Delta A), f(\Delta B))$ by $\gcd(a, b)$

Speed Analysis

Why is this method faster?

Getting the prime factorisation of small numbers is faster than large numbers. Dividing the LCM by the GCD makes the prime factorisation process faster.