I need help on this question. What is the smallest positive integer n such that 2n is a perfect square and 3n is a perfect cube?
I THINK THE ANSWER IS 1, IS THIS CORRECT?
Consider the prime factorization of n. How many factors of 2 does this need to contain and how many factors of 3?
prime factorization for 36
prime factorization for 180
To find the smallest positive integer n such that 2n is a perfect square and 3n is a perfect cube, we need to consider the prime factorization of n.
Let's start by assuming that n is written as n = 2^a * 3^b * k, where k is any positive integer and a and b are non-negative integers.
We know that 2n is a perfect square, which means that its prime factorization has only even exponents. So, the prime factorization of 2n can be written as 2^c, where c is an even integer.
Similarly, 3n should be a perfect cube, so its prime factorization has only multiples of 3 as exponents. Therefore, the prime factorization of 3n can be written as 3^d, where d is a multiple of 3.
Now, let's consider the prime factorization of 2n and 3n:
2n = 2^a * 3^b * k
3n = 2^c * 3^d
Since 2n is a perfect square, the exponents in the prime factorization of 2n must be even. Similarly, since 3n is a perfect cube, the exponents in the prime factorization of 3n must be multiples of 3.
We can derive some conditions from this:
1. The exponent of 2, a, must be even so that 2n is a perfect square.
2. The exponent of 3, b, must be a multiple of 3 so that 3n is a perfect cube.
To find the smallest positive integer n that satisfies these conditions, we need to find the smallest values of a and b that meet the requirements.
For a to be even, the smallest possible value is 0. For b to be a multiple of 3, the smallest possible value is 3.
So, if we set a = 0 and b = 3, the prime factorization of n becomes:
n = 2^0 * 3^3 = 1 * 3^3 = 27
Therefore, the smallest positive integer n that satisfies the given conditions is n = 27.
So, the answer is n = 27, which means your initial guess of 1 was incorrect.