If a and b are two distinct positive integers such that b > a, which

Asked: If a and b are two distinct positive integers such that b > a, which of the following expressions is equivalent to the difference between the maximum and the minimum possible values of the ratio of the least common multiple of a and b to the highest common factor of a and b ?

\(HCF_{min}(a,b) = 1\)
\(LCM_{max}(a,b)= ab\)
\(HCF_{max}(a,b) = a\)
\(LCM_{min}(a,b)= b\)

\(Max(\frac{LCM(a,b)}{HCF(a,b)}) = \frac{LCM_{max}(a,b) }{ HCF_{min}(a,b)} = \frac{ab }{ 1} = ab\)
\(Min (\frac{LCM(a,b)}{HCF(a,b)}) = \frac{LCM_{min}(a,b) }{ HCF_{max}(a,b)} = \frac{b}{a} \)
Difference between the maximum and the minimum possible values of the ratio of the least common multiple of a and b to the highest common factor of a and b = \(ab - \frac{b}{a} = \frac{b}{a} (a^2 - 1) = \frac{b}{a} (a+1)(a-1)\)

IMO C

We should try to focus on finding the ratio of LCM / HCF of exotic cases first. Think about the extreme cases.

Let HCF to be 1 so that \(a\) and \(b\) have no factors in common but 1. In that case, the LCM is \(a*b\). Then the ratio is \(ab/1 = ab\).

Let's also think about the extreme case where \(b\) is a multiple of \(a\). Then the HCF would be \(a\), LCM would be \(b\) since \(b\) is already a multiple of \(a\). Then the ratio would be \(b/a\).

Now that we listed the two extreme cases, we can see the first case is a maximum and the second would correspond to the minimum.

The difference is now \(ab - b/a = \frac{b}{a}*(a^2 - 1) = \frac{b}{a}*(a-1)(a+1)\)

Ans: C

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