25 has three factors: 1, 5, and 25. Their product is 

\(\displaystyle 1 \times 5 \times 25 = 125.\)

None of the choices are prime, so each has at least three factors. The question, then, is which one has only three factors?

We can eliminate four choices by showing that each has at least four factors - that is, at least two different factors other than 1 and itself:

\(\displaystyle 120 = 2 \times 60\)

\(\displaystyle 122 = 2 \times 61\)

\(\displaystyle 123 = 3 \times 41\)

\(\displaystyle 124 = 2 \times 62\)

Each, therefore, has at least four factors.

However, the only way to factor 121 other than \(\displaystyle 1 \times 121\) is \(\displaystyle 11 \times 11\). Therefore, 121 has only 1, 11, and 121 as factors, and it is the correct choice.

60 has twelve factors: 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, and 60.

Their sum is \(\displaystyle 1 + 2 + 3 + 4 + 5 + 6 + 10 + 12 + 15 + 20 + 30 + 60 = 168\).

\(\displaystyle 135 = 5 \times 27 = 5 \times 3 \times 9= 5 \times 3 \times 3 \times 3\)

3 and 5 are both primes, so this is as far as we can go. Rearranging, the prime factorization is 

\(\displaystyle 135 =3 \times 3 \times 3 \times 5\).

Place each of these digits into the box in turn. Divide each of the numbers formed and see which quotient yields a zero remainder:

\(\displaystyle 627 \div 3 = 209 \textrm{ R } 0\)

\(\displaystyle 637 \div 3 = 212 \textrm{ R } 1\)

\(\displaystyle 667 \div 3 = 222 \textrm{ R } 1\)

\(\displaystyle 677 \div 3 = 225 \textrm{ R } 2\)

\(\displaystyle 697 \div 3 = 232 \textrm{ R } 1\)

Only 627 is divisible by 3 so the correct choice is 2.

For a number to be divisible by 4, the last two digits must form an integer divisible by 4. 2 (02), 22, 62, and 82 all yield remainders of 2 when divided by 4, so none of these alternatives make the number a multiple of 4.

Numbers that are divisble by 6 are also divisble by 2 and 3. Only even numbers are divisible by 2, therefore, 72165 is excluded. The sum of the digits of numbers divisible by 3 are also divisible by 3. For example,

\(\displaystyle 6+3+0+7+2=18\)

Because 18 is divisible by 3, 63,072 is divisible by 3.

The elements are as follows:

\(\displaystyle A = \left \{ 5, 10, 15, 20...395 \right \}\)

This can be rewritten as

\(\displaystyle A = \left \{ 5 \times 1, 5\times 2, ... 5 \times 79 \right \}\).

Therefore, there are \(\displaystyle 79\) elements in \(\displaystyle A\).

The elements are as follows:

\(\displaystyle A = \left \{ 3, 6, 9, 12, 15...399 \right \}\)

This can be rewritten as

\(\displaystyle A = \left \{ 3 \times 1, 3 \times 2, ... 3 \times 133 \right \}\).

Therefore, there are \(\displaystyle 133\) elements in \(\displaystyle A\).

19 is a prime number and has 1 and 19 as its only factors. Their sum is 20.