Finding the possible Least Common Multiples of of numbers with Highest Common Factor 8

Question

The Highest Common Factor of two numbers is 8. Which one of the following can never be their Least Common Multiple?

The choices are as follow:

A. 8

B. 12

C. 60

D. 72

The answer key states that the answer is 60 because 8 is not a factor of 60. But 8 is not a factor of 12 either. If I remember it right, there’s this factor foundation rule wherein the factors of the factor of a certain integer also divides evenly into that integer. In which case, if 4 is also one factor of 8, then 60 can be divided evenly by 4. Hence, 60 can probably be an LCM of the HCF, 8.

Answer 1

You are correct that both 12 and 60 cannot be the least common multiple of two numbers whose highest common factor is 8. There's probably a typographical error in your text.

Answer 2

The answer key is correct that $60$ cannot be the LCM, but incorrect if it claims that $12$ can be the LCM. If the HCF of $a$ and $b$ is $8$, the LCM is $\frac {ab}8$ and since $8$ divides both $a$ and $b$ it divides the LCM.

Answer 3

The general rule is that $(d,m)$ of positive integers can be the pair of the highest common factor $\def\hcf{\operatorname{hcf}}\def\lcm{\operatorname{lcm}}d=\hcf(a,b)$ and lowest common multiple $m=\lcm(a,b)$ of two positive integers $a,b$ if and only if $d~$divides$~m$. The the condition is necessary is because $d$ must divide $a$ which must divide $m$, by definition, and divisibility is transitive. (One could also use $b$ as intermediate value.) The condition is sufficient because if $d~$divides$~m$, one can simply take $(a,b)=(d,m)$; clearly in that case one has $\hcf(d,m)=d$ and $\lcm(d,m)=m$.

Answer 4

The two numbers must be $8x$ and $8y$ where $\gcd(x,y) = 1$. Hence $\operatorname{lcm}(8x, 8y) = 8xy$. So we need to find relative prime integers $x$ and $y$ such that

 A. 8xy = 8
 B. 8xy = 12
 C. 8xy = 60
 D. 8xy = 72

Clearly (B) and (C) have no integer solutions.