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2. The number of odd divisors of any number, which is in the form of [tex]2^{k}[/tex] (where [tex]k \in \mathbb{N}[/tex])

A. 2
B. 1
C. 3
D. 4


Sagot :

Let's analyze the problem by understanding the characteristics of numbers in the form of [tex]\(2^k\)[/tex], where [tex]\(k\)[/tex] is an arbitrary integer.

Step 1: Express the Number

The number given is [tex]\(2^k\)[/tex]. For example:
- [tex]\(2^1 = 2\)[/tex]
- [tex]\(2^2 = 4\)[/tex]
- [tex]\(2^3 = 8\)[/tex]
- And so on.

Step 2: Identify Properties of [tex]\(2^k\)[/tex]

Numbers in the form [tex]\(2^k\)[/tex] are powers of 2. [tex]\(2^k\)[/tex] is constituted solely of the factor 2 and involves repeated multiplication of the number 2.

Step 3: Consider What an Odd Divisor Is

An odd divisor is a divisor that is not divisible by 2. In other words, it has no factor of 2.

Step 4: Determine Odd Divisors of [tex]\(2^k\)[/tex]

Since [tex]\(2^k\)[/tex] is made up entirely of the factor 2:
- It is an even number for all [tex]\(k \geq 1\)[/tex].
- The only possible odd number is 1.

Step 5: Conclusion

The only odd divisor of [tex]\(2^k\)[/tex] (since it only has the factor 2) is the number 1. This is because 1 is the only number that is a divisor of every integer and is odd.

Thus, the number of odd divisors of [tex]\(2^k\)[/tex] is:

Answer: 1

Therefore, the correct answer is:
b) 1