HCF and LCM given need to identify two numbers that correspond

[scon1]

How do I go about identifying two numbers with a LCM of 30 and and HCF of 3, in a simple way?

Thank you in advance.

 

[pka]

How do I go about identifying two numbers with a LCM of 30 and and HCF of 3, in a simple way?

I found two pairs of positive integers that work. Can you?

 

[scon1]

Positive integers

It is positive integers that I need, I can find them by listing all LCM and HCF involving the 30 and 3 but it seems very long winded. Google suggested factorising, which is fine as I can list the prime factors but I lose the method at the final stage when using the prime factors to derive the answer.

 

[stapel]

Google suggested factorising, which is fine as I can list the prime factors but I lose the method at the final stage when using the prime factors to derive the answer.

The first result in Google is this one. Have you tried listing things out in a table in the way the linked page displays? :wink:

 

[soroban]

Hello, scon1!

How do I go about identifying two numbers with a LCM of 30 and and HCF of 3, in a simple way?

Consider two positive integers \(\displaystyle a\) and \(\displaystyle b\) and their prime factorization.

We are told that: \(\displaystyle \begin{array}{c}LCM(a,b) = 30 \\ HCF(a,b) = 3 \end{array} \)

This means: .\(\displaystyle \begin{array}{ccc} \{\text{factors of }a\}\,\cup\,\{\text{factors of }b\} &=& \{2,3,5\} \\ \{\text{factors of }a\}\,\cap\,\{\text{factors of }b\} &=& \{3\} \end{array}\)

\(\displaystyle a\) and \(\displaystyle b\) both have a factor of 3,
. . and together they have factors of 2, 3, and 5.


Two solutions are possible:.\(\displaystyle \begin{array}{ccc}(2\!\cdot\!3,\,3\!\cdot\!5) \:=\: (6,15) \\ (3,\,2\!\cdot\!3\!\cdot\!5) \:=\: (3,30)\end{array}\)

 

[lookagain]

How do I go about identifying two numbers with a LCM of 30 and and HCF of 3, in a simple way?

Thank you in advance.

Be aware that the LCM(a, b)*HCF(a, b) = the product ab.