## Task 3 - 310623

How many natural numbers are there in total that are

a) divisors of 256,

b) divisors of $2 \cdot 256$,

c) divisors of $256 \cdot 256$?

Explain how you found each of your three answers!
a) One observes: 256 is divisible by 2; it holds that $256: 2=128$. Further, $128: 2=64$. By continuing to check divisibility by 2, we get

$$
256=2 \cdot 128=2 \cdot 2 \cdot 64=\ldots=2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2
$$

Thus, a natural number is a divisor of 256 if and only if it is either equal to 1 or equal to the product of a certain number of factors 2, where this number is at most 8.

For this number, there are exactly 8 possibilities. Therefore, there are a total of 9 natural numbers that are divisors of 256.

b) From $2 \cdot 256=2^{9}$;

c) from $256 \cdot 256=2^{16}$ it follows: There are a total of 10 and 17 natural numbers, respectively, that are divisors of $2 \cdot 256$ and $256 \cdot 256$.