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If $n$ and $k$ are positive integers such that $5<\frac nk<6$, then what is the smallest possible value of $\frac{\mathop{\text{lcm}}[n,k]}{\gcd(n,k)}$?

I am really not sure where to start. I know that in order to create the minimum value, $n$ and $k$ should share a common factor. However, I keep plugging in numbers to no avail. Help is greatly appreciated.

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First of all all fractions can be put into lowest terms.

Second of all If $\frac nk = \frac ab$ and $\gcd(n,k) = d$ and $gcd(a,b) = e$ and $a = a'e; b= b'e;n=n'd; k= k'd$, then $\frac ab =\frac nk \implies \frac {a'e}{b'e}=\frac{n'd}{k'd}\implies\frac {a'}{b'} = \frac {n'}{k'}$ and both are in lowest terms. So $a' = n'$ and $b' = k'$.

Third of all. $\frac{\mathop{\text{lcm}}[n,k]}{\gcd(n,k)} = \frac {n'k'd}{d} = {n'k'}$ and $\frac{\mathop{\text{lcm}}[a,b]}{\gcd(a,b)} = \frac {a'b'e}{e} = a'b' = {n'k'}$.

So we might as well assume $\frac nk$ is in lowest terms.

So we have $5 < \frac nk < 6$ and $n,k$ is in lowest terms and we want to find the least possible value of $nk$.

$5 < \frac nk < 6$

$5k < n < 6k$.

So $kn > k*(5k) = 5k^2$.

As there is no possible $n$ so that $5 < \frac n1 < 6$, the smallest possible value of $k$ is $2$ and if $5 < \frac n2 < 6$ we must have $n =11$. That is one possible solution.

In that case $nk = 11*2 = 22$. (And notice: $22 > 5*2^2$). Is that the smallest possible value?

If $k \ge 3$ we will have $kn >5k^2 \ge 5*3^2 =45 > 22$.

So, yes, $22$ is the smallest possible value.

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Without loss of generality, you can assume that $n$ and $k$ are coprime, hence that $\gcd(n,k)=1$, because this neither changes the value of $\frac{n}{k}$ nor of $\frac{\mathrm{lcm} (n,k)}{\gcd (n,k)}$. Note that this last fraction is indeed nothing else than the product of the "not-common factors" of $n$ and $k$ (in particular, an integer). Then your task is to minimize $\mathrm{lcm}(n,k)$. But now this is easy, because it is just $n\cdot k$. So what is the smallest possible value of $n \cdot k$ if you have the restriction that $\frac{n}{k} \in (5,6)$? Clearly $22$, by setting $n=11, k=2$. Indeed, $k$ must be at least $2$, hence $n$ must be at least $11$.

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