"Give them the gift of words"
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.