Find an integer n such that n/2 is a square, n/3 is a cube, and n/5 is a fifth power.
Is there a way to do this without using a computer? If so, can anyone give me a hint. It's killing me.
Hello, paupsers!
$\displaystyle \text{Find an integer }n\text{ so that }\dfrac{n}{2}\text{ is a square, }\dfrac{n}{3}\text{ is a cube, and }\dfrac{n}{5}\text{ is a fifth power.}$
Since $\displaystyle \,n$ is divided by 2, 3, and 5,
. . we assume that $\displaystyle \,n$ has factors of 2, 3, and 5 (at least).
Let $\displaystyle n \;=\;2^a\cdot 3^b\cdot5^c$ . for some positive integers $\displaystyle a,b,c.$
$\displaystyle \dfrac{n}{2}$ is a square.
We have: .$\displaystyle \dfrac{n}{2} \;=\;2^{a-1}\cdot 3^b\cdot 5^c$
. . where: .$\displaystyle a\!-\!1,\;b,\;c$ are all even. .[1]
$\displaystyle \dfrac{n}{3}$ is a cube.
We have: .$\displaystyle \dfrac{n}{3} \;=\;2^a\cdot3^{b-1}\cdot 5^c$
. . where: .$\displaystyle a,\;b\!-\!1,\;c$ are all mutiples of 3. .[2]
$\displaystyle \dfrac{n}{5}$ is a fifth power.
We have: .$\displaystyle \dfrac{n}{5} \;=\;2^a\cdot 3^b \cdot 5^{c-1}$
. . where: .$\displaystyle a,\;b,\,c\!-\!1$ are all multiples of 5. .[3]
Combining [1], [2] and [3], we have:
$\displaystyle \begin{Bmatrix}a\1-\!1\text{ is even} \\ a\text{ is a multiple of 3} \\ a\text{ is a multiple of 5} \end{Bmatrix}$ . This is satisfied by: .$\displaystyle a\,=\,15$
$\displaystyle \begin{Bmatrix}b\text{ is even} \\ b\!-\!1\text{ is a multiple of 3} \\ b\text{ is a multiple of 5}\end{Bmatrix}$ .This is satisfied by: .$\displaystyle b \,=\,10$
$\displaystyle \begin{Bmatrix}c\text{ is even} \\ c\text{ is a multiple of 3} \\ c\!-\!1\text{ is a multiple of 5}\end{Bmatrix}$ . This is satisfied by: .$\displaystyle c \,=\,6$
Therefore, the least positive integer $\displaystyle \,n$ is:
. . . . $\displaystyle n \;=\;2^{15}\!\cdot3^{10}\!\cdot5^6$