Let $x$ be the smallest positive integer that simultaneously satisfies the following conditions: $2 x$ is the square of an integer, $3 x$ is the cube of an integer, and $5 x$ is the fifth power of an integer. Find the prime factorization of $x$.

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Solution

From the statement, we can write

$$
2 x=a^{2}, 3 x=b^{3}, 5 x=c^{5}
$$

where $a, b, c$ are integers. Therefore, we can conclude that $x$ has prime factors 2, 3, and 5. That is, the prime factorization can be written as $x=2^{p} 3^{q} 5^{r} \cdot m$, where $m$ is the product of primes greater than 5. Due to (1), $m$ must be a perfect square, a perfect cube, and a perfect fifth power. Since we want the smallest possible $x$, we can assume that $m=1$.

In general, the prime factorization of a power $n^{k}$ can be obtained from the prime factorization of $n$ by multiplying each exponent of its prime factors by $k$. Therefore, the exponents in the prime factorization of a perfect square are all even, those of a perfect cube are all multiples of 3, and those of a perfect fifth power are all multiples of 5. Thus, equation (1) tells us that
i) $p+1, q$, and $r$ are multiples of 2;

ii) $p, q+1$, and $r$ are multiples of 3;

iii) $p, q$, and $r+1$ are multiples of 5.

From i) and ii), the smallest possible value for $p$ is $3 \cdot 5=15$. Note that this value satisfies i). From i) and ii), the possible values for $q$ are $10, 20, 30, \ldots$ The first of these does not satisfy i), so the smallest possible value for $q$ is 20. Finally, from i) and iii), the possible values for $r$ are $6, 12, 18, 24, \ldots$. From this list, the smallest value that satisfies iii) is 24. Therefore, the smallest value of $x$ is $2^{15} 3^{20} 5^{24}$.