Prime Numbers/Sum/Odd/Even | Number Properties

Number properties concepts come across as pretty easy, theoretically, but they have some of the toughest questions. Today let’s take a look at some properties of prime numbers and their sum. Note that don’t try to “learn” all the takeaways you come across for number properties – it will be very stressful. Instead, try to understand why the properties are such so that if you get a question related to some such properties, you can replicate the results effortlessly.

To start off, we would like to take up a simple question and then using the takeaway derived from it, we will look at a harder problem.

Question 1: Which of the following CANNOT be the sum of two prime numbers?

(A) 19

(B) 45

(D) 79

(E) 88

Solution: What do we know about sum of two prime numbers?

e.g. 3 + 5 = 8

5 + 11 = 16

5 + 17 = 22

23 + 41 = 64

Do you notice something? The sum is even in all these cases. Why? Because most prime numbers are odd. When we add two odd numbers, we get an even sum.

We have only 1 even prime number and that is 2. Hence to obtain an odd sum, one number must be 2 and the other must be odd.

2 + 3 = 5

2 + 7 = 9

2 + 17 = 19

Look at the options given in the question. Three of them are odd which means they must be of the form 2 + Another Prime Number.

Let’s check the odd options first:

(A) 19 = 2 + 17 (Both Prime. Can be written as sum of two prime numbers.)

(B) 45 = 2 + 43 (Both Prime. Can be written as sum of two prime numbers.)

(D) 79 = 2 + 77 (77 is not prime.)

79 cannot be written as sum of two prime numbers. Note that 79 cannot be written as sum of two primes in any other way. One prime number has to be 2 to get a sum of 79. Hence there is no way in which we can obtain 79 by adding two prime numbers.

(D) is the answer.

Now think what happens if instead of 79, we have 81?

81 = 2 + 79

Both numbers are prime hence all three odd options can be written as sum of two prime numbers. Then we would have had to check the even options too (at least one of which would be different from the given even options). Think, how would we find which even numbers can be written as sum of two primes? We will give the solution of that next week. So the takeaway here is that if you get an odd sum on adding two prime numbers, one of the numbers must be 2.

Question 2: If ${m}$ , ${n}$ and ${p}$ are positive integers such that ${m < n < p}$ , is ${m}$ a factor of the odd integer ${p}$ ?

Statement 1: ${m}$ and ${n}$ are prime numbers such that ${ (m + n)}$ is a factor of 119.

Statement 2: ${p}$ is a factor of 119.

Solution: First of all, we are dealing with positive integers here – good. No negative numbers, 0 or fraction complications. Let’s move on.

The question stem tells us that $p$ is an odd integer. Also, $m < n < p$ .

Question: Is $m$ a factor of $p$ ?

There isn’t much information in the question stem for us to process so let’s jump on to the statements directly.

Statement 1: $m$ and $n$ are prime numbers such that $(m + n)$ is a factor of 119.

Write down the factors of 119 first to get the feasible range of values.

119 = 1, 7, 17, 119

All factors of 119 are odd numbers. So $(m + n)$ , a sum of two primes must be odd. This means one of $m$ and $n$ is 2. There are many possible values of $m$ and $n$ e.g. 2 and 5 (to give sum 7) or 2 and 15 (to give sum 17) or 2 and 117 (to give sum 119).

Note that we also have $m < n$ . This means that in each case, $m$ must be 2 and $n$ must be the other number of the pair.

So now we know that $m$ is 2. We also know that $p$ is an odd integer. Is $m$ a factor of $p$ ? No. Odd integers are those which do not have 2 as a factor. Since $m$ is 2, $p$ does not have $m$ as a factor.

This statement alone is sufficient to answer the question!

Statement 2: $p$ is a factor of 119

This tells us that $p$ is one of 7, 17 and 119. $p$ cannot be 1 because $m < n < p$ where all are positive integers.

But it tells us nothing about $m$ . All we know is that it is less than $p$ . For example, if $p$ is 7, $m$ could be 1 and hence a factor of $p$ or it could be 5 and not a factor of $p$ . Hence this statement alone is not sufficient.

Answer (A).

Something to think about: In this question, if you are given that $m$ is not 1, does it change our answer?

Key Takeaways:

– When two distinct prime numbers are added, their sum is usually even. If their sum is odd, one of the numbers must be 2.

– Think what happens in case you add three distinct prime numbers. The sum will be usually odd. In case the sum is even, one number must be 2.

– Remember the special position 2 occupies among prime numbers – it is the only even prime number.

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