Q:

Find a positive integer such that half of it is a square,a third of it is a cube,and a fifth of it is a fifth power

I found this problem in Underwood Dudley book in Linear congruence and Chineese remainder theorem ,but I can't present it as system of Linear congruence Anyone have any idea?
plz help me

A:

I bet this is not what you were looking for, but it is worth mentioning that $0$ is a solution.

A:

Guess that the number is of the form $2^{n_2} 3^{n_3} 5^{n_5}$. Then
\begin{align}
n_2-1 \equiv n_3 \equiv n_5 \equiv 0 \mod 2\\
n_2 \equiv n_3-1 \equiv n_5 \equiv 0 \mod 3\\
n_2 \equiv n_3 \equiv n_5-1 \equiv 0 \mod 5
\end{align}
Rewriting this, we have
\begin{align}
n_2 \equiv 0 \mod 15,\quad n_2 \equiv 1 \mod 2\\
n_3 \equiv 0 \mod 10, \quad n_3 \equiv 1 \mod 3\\
n_5 \equiv 0 \mod 6, \quad n_5 \equiv 1 \mod 5
\end{align}
Some trial and error can give you $n_2 = 15$, $n_3=10$, and $n_5=6$ as a simple solution (among many others). So $2^{15} 3^{10} 5^6$ is one solution.

A:

A positive integer works if and only if it is of the form $2^a3^b5^cn^{30}$, and the following must hold:
$a\equiv 1 \bmod 2, 15|a$.
$b\equiv 1 \bmod 3,10|b$.
$c\equiv 1 \bmod 5,6|c$.
I found them by trial and error, but there are methods to solve them, (look up chinese remainder theorem.)
We get solutions: $a=15,b=10,c=6$
So $2^{15}3^{10}5^{6}$ works.

