## Answers

**(a)** Obviously and for a prime number , since its only divisors are and itself.

1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |

1 | 2 | 2 | 3 | 2 | 4 | 2 | 4 | 3 | 4 |

For the other values one checks it manually,

**(b)** First note that . All the divisors of are of the form for , and . So, the positive divisors of are in correspondence with how many combinations of exponents we can choose. In this case, .

**(c)** All the divisors of are . Therefore, .

**(d)** The function is not one-to-one.

As we saw in *(a)*, . That is, two different values have the same output.

**(e)** The function is onto. Indeed, for we have as we saw in *(c)*.

**(f)** The set is the set of prime numbers. Note that for we have that since and are always divisors. If it means that does not have any other divisor, hence it is prime.

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f(p) = 2

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90 = 2 x 32 x 5

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