If x is an integer greater than 1, is x equal to \(2^k\) for some positive integer k?If x is an integer greater than 1, is x equal to 2^k for some positive integer k?
(1) x has only one prime factor.
(2) Every factor of x is even.
Can anyone please explain the statements to me?
Statement 1 says x has only one prime factor. Am I to assume that x is a prime number? or am i to assume that x is a number such as 9, whose only prime factor is 3. Or the number can be 2,3... its confusing to me,
Statement 2 says every factor of x is even. Does such a number exist?
Source: Jeff Sackmann's questions
Basically question ask whether x is some power of 2: 2 (for k=1), 4 (for k=2), 8 (for k=3), ...
(1) x has only one prime factor --> x can be ANY prime in ANY positive integer power: 2, 2^3, 3, 3^7, 5, 5^2, ... Note that all this numbers have only one prime factor. Not sufficient.
(2) Every factor of x is even --> this statement makes no sense, every positive integer has at least one positive odd factor: 1. I think it should be: x has no odd factor more than 1 (or: every factor of x, except 1, is even). In this case as x don't have any odd factors >1 then x has no odd primes in its prime factorization --> x is of the form of 2^k for some positive integer k. Sufficient.
Not a good question.
Yes. if statement 2 is rephrased as you said, it will be a sufficient statement. I think this what even the question maker had in mind but it slipped off his mind somehow. I will mail him to correct this.