1.3: Problem 16: Find a positive integer \(n\) such that \(n/2\) is a square, \(n/3\) is a cube, and \(n/5\) is a fifth power.

Solution

We want to construct a number that is divisible by 2, 3 and 5. Therefore, we need this number to be of the form

$$ \begin{align*} n = 2^a \cdot 3^b \cdot 5^c \end{align*} $$

Next, we want \(n/2\) to be a square, this means that we need each exponent to be even in

$$ \begin{align*} n/2 = 2^{a-1} \cdot 3^b \cdot 5^c \end{align*} $$

Similarly, we want \(n/3\) to be a cube. This means that each exponent is divisible by 3 in

$$ \begin{align*} n/3 = 2^a \cdot 3^{b-1} \cdot 5^c \end{align*} $$

Finally, we want \(n/5\) to be a fifth power. This means that each exponent is divisible by 5 in

$$ \begin{align*} n/5 = 2^a \cdot 3^b \cdot 5^{c-1} \end{align*} $$

So let’s analyze the constraints on each exponent. For \(a\), we want \(a - 1\) to be even so \(a\) must be odd. \(a\) must also be divisible by 3 and by 5. The smallest number satisfying this is \(a = 15\). For \(b\), we want \(b\) to be even and divisible by 5. We also need \(b-1\) to be divisible by 3. The smallest integer satisfying these conditions is \(b = 10\). Finally, \(c-1\) needs to be divisible by 5. \(c\) is even and divisible by 3. 6 is divisible by 3 and even and 6-1 is divisible by 5. Therefore, we have

$$ \begin{align*} n = 2^{15} \cdot 3^{10} \cdot 5^{6} = 30,233,088,000,000. \end{align*} $$

References