The square of a certain whole number N is N squared.If 60 is a factor of N squared ,its possible that xis not a factor of N squared.
true, depending on the value of x.
A.16
B.25
C.36
D.100
These are the choices for the question
60 divides N^2
60 = 2^2*3*5
That means that 3^2 and 5^2 must also divide N^2
So, only 16 might not be a factor of N^2
(2*3*5)^2 = 900
16 is not a factor of 900; then others are.
To determine if it is possible for 60 to be a factor of N squared while x is not a factor of N squared, we need to consider the prime factorization of 60 and understand the relationship between the prime factors of 60 and any potential factor x of N squared.
The prime factorization of 60 is: 2^2 x 3 x 5
Now, let's consider the prime factors of N squared. Since N is a whole number, N squared will have the same prime factors as N, each raised to the power of 2.
Let's assume that x is a factor of N squared. This means x must divide N squared evenly.
To determine if it is possible for 60 to be a factor of N squared while x is not a factor of N squared, we need to check if the prime factors of x are among the prime factors of N squared. If any prime factor of x is not present in the prime factorization of N squared, then x will not be a factor of N squared.
Since the prime factors of 60 are 2^2, 3, and 5, for x to be a factor of N squared, x must have these prime factors in its prime factorization.
Therefore, if x = 2^a x 3^b x 5^c (where a, b, and c are non-negative integers), for x to be a factor of N squared, the exponents a, b, and c must be even numbers. This ensures that every prime factor of x is present in the prime factorization of N squared, as each prime factor of N squared is already raised to the power of 2.
However, if any of the exponents a, b, or c are odd numbers (not even), then x will not be a factor of N squared.
So, it is possible that 60 is a factor of N squared while x is not a factor of N squared if x has prime factors that are not present in the prime factorization of N squared.