Final answer:
To find an integer n such that n/2 is a square, n/3 is a cube, and n/5 is a fifth power, we need n to have the prime factors of 2, 3, and 5. The integer n that satisfies these conditions is 360.
Step-by-step explanation:
To find an integer n such that n/2 is a square, n/3 is a cube, and n/5 is a fifth power, we can use the fact that a number raised to a certain power is equal to itself multiplied by itself that many times. Let's use variables to represent the powers: n/2 = x^2, n/3 = y^3, and n/5 = z^5. We want to find an integer n that satisfies these equations.
We can rewrite the equations as n = 2x^2, n = 3y^3, and n = 5z^5. Since n is the same in all three equations, we can set the right sides of the equations equal to each other: 2x^2 = 3y^3 = 5z^5. Now let's think about the prime factors of 2, 3, and 5.
2 only has the prime factor 2, 3 only has the prime factor 3, and 5 only has the prime factor 5. Therefore, we need n to have the prime factors of 2, 3, and 5. To find an integer n that satisfies this, we can take the product of 2, 3, and 5 raised to the highest power that appears in the equations: n = 2^3 * 3^2 * 5^2 = 360. So, the integer n that satisfies the given conditions is 360.