find all natural numbers n which are not divisible by 3 such that the number 2^n^2/2^10+2133 is a perfect cube.
视频信息
答案文本
视频字幕
We need to find all natural numbers n that are not divisible by 3, such that the fraction 2 to the power n squared, divided by 2 to the power 10 plus 2133, is a perfect cube. Let's start by analyzing this expression step by step.
First, let's calculate the denominator of our expression. We have 2 to the power 10 plus 2133. 2 to the power 10 equals 1024, so 1024 plus 2133 equals 3157. Therefore, our expression becomes 2 to the power n squared divided by 3157.
Next, we need to determine if 3157 is prime. We test divisibility by prime numbers starting with 2, 3, 5, and so on. After checking, we find that 3157 is indeed a prime number. Meanwhile, the numerator 2 to the power n squared has only one prime factor: 2.
For the fraction to be an integer, 3157 must divide 2 to the power n squared. However, since 3157 is prime and not equal to 2, it cannot divide any power of 2. Therefore, the fraction is never an integer for any natural number n.
Since the expression is never an integer, it cannot be a perfect cube. A perfect cube must be an integer, but our fraction is never an integer for any natural number n. Therefore, there are no natural numbers n which are not divisible by 3 such that the given expression is a perfect cube. The answer is that no such natural numbers exist.