Problems involving prime and composite numbers are often not too hard. It’s good to know prime numbers at least till 50, or even better, 100.

As a recap, the prime numbers till 50 are:
2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47.

Q. Amy is given a number x. She picks two numbers a and b and then generates a sequence of numbers x, x +a, x + a + b, x + 2a + b, x + 2a + 2b, i.e. she alternately adds a to get a new number and then b to get another number. Which two numbers should she choose for a and b so that she is guaranteed to generate a prime number, regardless of the choice of x?

a.) 2, 3
b) 3, 6
c) 2, 4
d) 7, 14

So we don’t want to get stuck with composite numbers alone. This happens if aand bhave a common factor. For example if a = 3 and b = 6, then picking x = 3 leaves us with composite numbers alone. So option a is the correct answer.

Q. There are k students in a class. What could be the value of k if every attempt to sort them perfectly into teams of greater than 2, but less than k fails?

a) 6
b) 17
c) 18
d) 301

k is clearly a prime. So the answer is 17.

This was a really easy question. A slightly hard question would be something like this:

Q. Which of the following could be a value of k if k has an odd number of factors?

a) 72
b) 36
c) 45
d) 60

You can factorise each number above. But a useful trick is to observe that factors come in pairs. So if a is a factor of n, so is n/a. So factors will be counted in pairs, unless we have a repeated factor, i.e. a factor a such that a=n/a. Or we need a2=n i.e. n should be a perfect square. So the answer is b.

This is a useful fact. In the previous question, the answer choices were small. But imagine if the same question were rephrased as follows.

Q. Which of the following could be a value of k if k has an odd number of factors?

a) 100000000
b) 2210269
c) 373176
d) 1264845

The answer is a because 100000000 is a perfect square.