# Thread: Is this easily solvable without a computer?

1. ## Is this easily solvable without a computer?

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.

2. Originally Posted by paupsers
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.

Write $n=p_1^{a_1}\cdot\ldots\cdot p_k^{a_k}\,,\,\,p_i\,\,prime\,,\,\,0 , then $n$ is

an m-th power iff $\forall i=1,2,...,k\,,\,\,a_i=0\!\!\pmod m$ .

Tonio

3. Could you elaborate more? Based on what you're saying, I need some product of primes where each prime is raised to the power 30? (Since 30 mod 2 = 30 mod 3 = 30 mod 2 = 0)

I'm still not seeing how to solve for all 3 of the conditions at one time...

4. Hello, paupsers!

$\text{Find an integer }n\text{ so that }\dfrac{n}{2}\text{ is a square, }\dfrac{n}{3}\text{ is a cube, and }\dfrac{n}{5}\text{ is a fifth power.}$

Since $\,n$ is divided by 2, 3, and 5,
. . we assume that $\,n$ has factors of 2, 3, and 5 (at least).

Let $n \;=\;2^a\cdot 3^b\cdot5^c$ . for some positive integers $a,b,c.$

$\dfrac{n}{2}$ is a square.
We have: . $\dfrac{n}{2} \;=\;2^{a-1}\cdot 3^b\cdot 5^c$

. . where: . $a\!-\!1,\;b,\;c$ are all even. .[1]

$\dfrac{n}{3}$ is a cube.
We have: . $\dfrac{n}{3} \;=\;2^a\cdot3^{b-1}\cdot 5^c$

. . where: . $a,\;b\!-\!1,\;c$ are all mutiples of 3. .[2]

$\dfrac{n}{5}$ is a fifth power.
We have: . $\dfrac{n}{5} \;=\;2^a\cdot 3^b \cdot 5^{c-1}$

. . where: . $a,\;b,\,c\!-\!1$ are all multiples of 5. .[3]

Combining [1], [2] and [3], we have:

$\begin{Bmatrix}a\1-\!1\text{ is even} \\ a\text{ is a multiple of 3} \\ a\text{ is a multiple of 5} \end{Bmatrix}$ . This is satisfied by: . $a\,=\,15$

$\begin{Bmatrix}b\text{ is even} \\ b\!-\!1\text{ is a multiple of 3} \\ b\text{ is a multiple of 5}\end{Bmatrix}$ .This is satisfied by: . $b \,=\,10$

$\begin{Bmatrix}c\text{ is even} \\ c\text{ is a multiple of 3} \\ c\!-\!1\text{ is a multiple of 5}\end{Bmatrix}$ . This is satisfied by: . $c \,=\,6$

Therefore, the least positive integer $\,n$ is:

. . . . $n \;=\;2^{15}\!\cdot3^{10}\!\cdot5^6$