Brenda's favourite number

Assume the number is divisible by 3. Then either the number is 3 itself which is a prime or a composite number. For the former, we see a problem with Eric's statements. 3 is not greater than 30 and 3 has only two factors and two is non-composite which means both of Eric's statements are false. For the latter, we see a problem with Alfred's statements. A composite number is never prime and it is impossible for a multiple of 3 to have a digit sum of 7 (a non-multiple of 3). So both of Alfred's statements are false.

This means the mystery number cannot be divisible by 3. So statement 1 made by Brenda is false which means her statement 2 must be true. The number must have at least an even digit. Looking at Charles' statements, we see that a non-multiple of 3 is definitely not divisible by 6 since 6 is divisible by 3. Since Charles' statement 1 is false, his statement 2 is true. This means the number is at most 100.

Now assume the digit sum is 7. We have 7, 16, 25, 34, 43, 52, 61 and 70 as possibilities. This means the number cannot be prime. Hence we eliminate 7, 43 and 61. If the number is 16 or 25, we have a problem with Darius' statements as these are square numbers whose digits differ by a prime number. If the number is 34, 52 or 70, we have a problem with Eric's statements as these numbers are greater than 30 and each have a composite number of factors. Hence the number cannot have a digit sum of 7.

Since Alfred's statement 2 is false, his statement 1 is true. So the number is prime. This invalidates Darius' statement 1 as primes are never squares. This means the digits of the number differs by a prime. Looking at Alfred's statements, the number doesn't have a digit sum of 7. Considering Eric's statements, a prime has two factors which invalidates his statement 2. Hence the number is greater than 30. By elimination, the only prime which satisfies all the descriptions is 47.

Hence Brenda likes the number $\boxed{47}$ !!!

Darius's and Alfred's first statements are contradictory, so the Brenda's number is either one of these two possibilities :
1) a perfect square with digit sum of 7, or
2) a prime number with its units digit exceeds its tens digit by a prime number.

Now, Charles said something about the number being a multiple of 6 in his first statement, but with 6 being a composite, then it's impossible for Charles' first statement to belong within the second possibility, so Brenda's number is characterized further as :
1) a perfect square divisible by 6 with digit sum of 7 (implying even parity), or
2) a prime number less than 100 with its units digit exceeds its tens digit by a prime number (implying a 2 digit odd prime number).

With the two possibilities above, we can see that within 1) there is a contradiction between its own clues. If a perfect square is divisible by 6, then it's also divisible by 3 since 6 = 2 × 3, and furthermore, it's also divisible by 9 since there must be at least double the quantity of 3 as a factor of a perfect square (3² = 9), and because of that, the digit sum is always 9 and not 7.

We now know that the answer must be from the second possibility, so let's look for more clues from Eric and Brenda herself. Since the number is prime, the factors are always two (and those are 1 and itself), while 2 itself is a prime number and and therefore NOT a composite. Thus, Eric's second statement is a lie and his first should be our fourth clue.

Brenda's divisibility by 3 can only work without the implied 2 digit from Darius's second statement, so this is the lie and Brenda's clue is, "There is at least one even digit in the number."

In conclusion, her number must be "a prime number (by Alfred) between 30 (by Eric) and 100 (by Charles) in which its units digit exceeds its tens digit by a prime number (by Darius) with at least one even digit in the number (by Brenda)". It could be 47 (prime different of 3), 49 (prime different of 5) or 69 (prime different of 3), but 47 is the only true prime number here, and thus the only solution is { 47 }.