Find the smallest possible integer $y$ such that $288y$ is a multiple of $5616$

Question

Find the smallest possible integer $y$ such that $288y$ is a multiple of $5616$

$$288 = 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 3 \cdot 3$$

$$5616 = 2 \cdot 2 \cdot 2 \cdot 2 \cdot 3 \cdot 3 \cdot 3 \cdot 13$$

Answer 1

$288 = 2^5 \times 3^2$, and $5616=2^4 \times 3^3 \times 13$.

Therefore, we have that the LCM is $2^5 \times 3^3 \times 13$.

Dividng by $288$ to get $y = 3 \times 13=\boxed{39}$

Answer 2

We want $5616 \mid 288 y$. This is possible if $288y$ contains all the prime factors from $5616$.

Using your factorizations (I did not check them): \begin{align} 288 &= 2\cdot2\cdot2\cdot2\cdot2\cdot3\cdot3\\ 5616 &= 2\cdot2\cdot2\cdot2\cdot3\cdot3\cdot3\cdot13 \end{align} and realigning a bit \begin{align} 288 &= 2\cdot2\cdot2\cdot2\cdot2\cdot3\cdot3\\ 5616 &= 2\cdot2\cdot2\cdot2\cdot1\cdot3\cdot3\cdot3\cdot13 \end{align} we see that we just lack the factors $3$ and $13$ in the factors of $288$ to have all the needed factors for $5616$.

Thus $y = 3 \cdot 13$. We used no uneeded factors, so this number is minimal.

Answer 3

We want $5616n=288y$, where $n,y\in N$ and $y\to min.$

Reducing we get $y=\frac{39n}{2}$. Since $y$ and $n$ are proportional, min $y$ means min $n=2$, consequently $y=39$.