How to find the unknown number while only LCM is given

Question

Find the smallest value of $n$ such that the lcm of $n$ and $15$ is $45$. I have searched many ways to solve but could not find a step wise method to teach the students, please help.

Answer 1

I would present the following as a systematic approach $$ \eqalign{ & {\rm lcm}(15,n) = 45 = {{15 \cdot n} \over {\gcd (15,n)}}\quad \Rightarrow \quad n = 3\gcd (3 \cdot 5,n)\quad \Rightarrow \cr & \Rightarrow \quad \min (n) = 3\min \left( {3,5,15} \right) = 9 \cr} $$

Answer 2

$n$ must be a divisor of $45$, however $n$ and $15$ must not be each other's divisor. Divisors of $45$: $$1,3,5,9,15,45.$$ Note that $3,5,15$ are divisors of $15$. So $n$ can be $9$ or $45$. Therefore the smallest $n$ is $9$.

Answer 3

$15=3\times 5$ and $45=3^2\times 5$. Your number $n=3^2=9$.

Answer 4

Hint:   $\,n \mid 45 = 3^2 \cdot 5\,$ so $\,n=3^a \cdot 5^b\,$ with $\,0 \le a \le 2\,$ and $\,0 \le b \le 1\,$. Find the smallest $\,a,b\,$ such that $\,45 = \operatorname{lcm}(15,n) = \operatorname{lcm}(3 \cdot 5, 3^a \cdot 5^b) = 3^{\max(1, a)}\cdot 5^{\max(1,b)}\,$.

Answer 5

We know that the Highest Common Factor is $45$, and that one of the numbers is $15$, and the other number is $n$

GCD=$\frac{45}{15}=3$.

L.C.M × H.C.F = $15\cdot n$

$3 × 45 = 15 × n $

$\frac{3 × 45}{15} = n$

$9 = n$