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GMAT Preparation Quant

3 Important Properties of Prime Numbers

Prime number is a key topic in GMAT Quant. You will get numerous questions based on Prime numbers in GMAT, so if you want to do well in the test, being thorough with Prime numbers is important.  In this article, we will discuss the application of 3 important properties of Prime numbers that can help you gain accuracy in GMAT questions. These properties are:

  • 2:  the only even prime number
  • 2 & 3: only primes which are consecutive integers
  • Prime number’s representation: (6n + 1) and (6n -1) – when and how to use.
Importance of Prime Numbers in Gmat Quant

Let’s discuss these properties in detail and then solve a few questions to further solidify the concepts and their application –

2: The Only Even Prime Number

The Only Even Prime Number - 2!
The Only Even Prime Number!

Before we discuss the unique properties of the number 2, we need to first understand – What is a Prime Number?

Primes numbers are natural numbers that are greater than 1 and have exactly two distinct factors – where one factor is 1 and the other factor is the number itself.

Keeping the above definition in mind, the following numbers can be regarded as Prime:

2, 3, 5, 7, 11…… and so on.

Notice that the smallest natural number which satisfies the definition of a prime number is 2 which is even and the remaining prime numbers are odd.

Wondering how we know that there aren’t any other even numbers? Well, that’s a valid question. To answer this question, let’s understand what is an even number?

An even number is of the form = 2*n, where n is an integer.

So, any even number(except when n = 1), will have minimum 1, 2, n, and 2n as its factors, which means it will have more than 2 factors. And we know that a prime number should have exactly 2 factors. Thus, we can infer that “2 is the only even prime number”.

This unique characteristic of 2 is tested amply in GMAT, especially in data sufficiency problems. So, it is important to remember the fact that “2 is the only even prime number” and if you are solving a DS question based on the prime numbers by plugging different values, then you must make sure to consider 2 as well.

There are several scenarios where this information comes handy. Examples of few such scenarios are given below:

Example 1: (Scenario 1- In DS question)

Integer n is greater than 1. Is n a prime number?

  1. n is an even number.
  2. Number n has no factor between 1 and n.

Solution

Step 1: Analyse Question Stem
  • n is an integer and n > 1.

We need to find out if n is prime.

Step 2: Analyse Statements Independently (and eliminate options) – AD/BCE

Statement 1: n an even number

  • n can be 2,4,6,8…..and so on. So, there will be two cases:
    • Case 1: If n = 2, then it will be a prime number.
    • Case 2: If n is an even number other than 2, then n is not a prime number.

Since, we are getting two contradictory results, we can infer that statement 1 is not sufficient and we can eliminate answer options A and D

Statement 2: Number n has no factor between 1 and n.

  • Since n has no factor between 1 and n and n is greater than 1, we can infer that –
    • The only factors of n will be 1 and n.
      • And since it has only 2 factors, we can infer that n is a prime number.

Hence, statement 2 is sufficient.

Thus, the correct answer is Option B.

Key learning from the solution

  • While solving the above example, we saw that Statement 1 was not sufficient. On the other hand, if a student forgets the fact that “2 is a prime number and it is even”, on seeing the Statement 1, they may think that n is an even number and has more than 2 factors. This will lead them to believe that n will never be a prime number and they may end up marking option A, which is incorrect.
  • If a student doesn’t know the definition properly, they may end up ignoring option B, without understanding that it’s actually telling us-that the number is prime.

Remember: An even number can be prime if that even number is 2.

Now, let’s take up another scenario.

Example 2 (Scenario 2- Even – odd nature of a number)

 If p, q, and r are 3 distinct positive integers such that (p +q) = k + r,  where k is an even integer, then is r an odd number?

  1. p, q, and r are prime numbers.
  2. r is a factor of 16.

Solution –

Step 1: Analyse Question Stem
  • p, q, and r are 3 distinct positive integers.
  •  p + q = k * r where r is an even integer.
    • even = even * r
    • So, r can be either even or odd.

We need to find if r is odd.

Step 2: Analyse Statements Independently (and eliminate options) – AD/BCE

Statement 1: p, q, and r are prime numbers.

  • Since p + q = even it means p and q must be odd prime numbers (i.e. p is not equal to 2 and q is not equal to 2)
  • even = even * r . Since r is also a prime number, there will be two cases:
    • Case 1: If r = 2, then it is even
    • Case 2: If r is any other prime number such that r \neq 2 , then it will be odd.

The results of the above two cases are contradictory.

Hence, statement 1 is not sufficient and we can eliminate answer options A and D

Statement 2: r is a factor of 16.

  • Positive factors of 16 = 1, 2, 4, 8,16
    • Therefore, if r = 1, then it will be an odd number,
    • else if r = 2, 4, 8 or 16, then it will be an even number.
  • Again, the results of the above two cases are contradictory.

Hence, statement 2 is also not sufficient and we can eliminate answer Option B.

Step 3: Analyse Statements by combining.
  • From statement 1: r is a prime number.
  • From statement 2: r can be 1, 2, 4, 8 or 16
  • On combining both the statements we get, r = 2, which is an even number.

Thus, the correct answer is Option C.

Key learning from the solution

  • In the above example, we were asked to find the even-odd nature of r. Statement 1 says that r is a prime number, so some students may miss the fact that r can be an even number too (i.e. r=2), hence, they will incorrectly conclude that option A is correct.
  • Although this is not related to primes, however, while analysing the 2nd statement, some students may forget about 1 and may simply think that r is an even number. So, they may end up marking the answer as B or D too.

Remember that: “Not all prime numbers are odd.”

Now, let’s focus on the final scenario –

Example 3: (Scenario 3: Finding unit digit in the multiplication of prime numbers)

P is the product of the first 100 prime numbers. What is the unit’s digit of P^{253} ?

A. 0
B. 2
C. 3
D. 5
E. 9

Solution –

Step 1: Understand Question Statement
  • P = product of the first 100 prime numbers = 2*3*5*7*11……….so on.
  • We need to find the unit digit of P^{253}
Step 2: Define Methodology
  • P = 2*5*(3*7*11*13…. and so on) = 10*(3*7*11*13…. and so on)
    • Hence,the unit’s digit of P = 0
  • Now, we will use the cyclicity of 0 to find the unit’s digit of P^{253}
Step 3: Calculate the final answer
  • The cyclicity of 0 is 1. For example:
    • 10^1  = 10 ; unit’s digit = 0
    • 10^2 = 100 ; unit’s digit = 0
    • 10^3 = 1000 ; unit’s digit = 0, and so on.
    • Hence, irrespective of the value of m, if the unit’s digit of a number n is 0, then the unit’s digit of any of its power (i.e. n^m ) is 0.
      • Here, n and m are positive integers.
  • Therefore, the unit’s digit of P^{253} = 0

Thus, the correct answer is Option A.

Key learning from the solution

  • In the above example, some students may forget the fact that 2 is also a prime number then they will think that P must be an odd multiple of 5. So, the unit’s digit of P will be 5 and as cyclicity of 5 is 1, hence, the unit’s digit of P^{253} = 5 . They will end up marking option D which is incorrect.

So, always keep in mind to consider 2, not only in the questions based on the even-odd nature of Primes but also in questions based on the unit’s digit involving primes, too.

2 & 3:  Only Primes Which Are Consecutive Integers

2 and 3 are the only prime numbers that are consecutive integers
The Only Consecutive Primes

In the first section, we learned the basic definition of a Prime number and that 2 is the only even Prime number.

In this section, we will move a step ahead and will learn about the consecutive Prime numbers in addition to Prime numbers which are consecutive integers.

When we say consecutive integers, we mean integers that come one after the other. For example: -1 and 0, 0 and 1, 1 and 2, etc. Similarly, when we say consecutive Prime numbers, we mean the Prime numbers which come one after the other. For example, 2 and 3, 3 and 5, 5 and 7, 13 and 19, etc.

Now, notice an interesting pattern here:

2 and 3 are the only Prime numbers that are also consecutive integers.

The rest of the Prime numbers such as (3, 5) (5, 7) (7, 11) and so on are simply consecutive Prime numbers, but none of them are consecutive integers. This observation may seem trivial, however, it can prove to be useful in a few GMAT questions.

Let’s solve two problems based on this concept to have a better insight –

Example 1: (Scenario 1: Even-odd type question)

If p and q are Prime numbers that are consecutive integers and p < q, then which of the following must be true?

  1.  {(7p+9q)}^3 + 5q^2 - 36pq is even.
  2.  (105p +280q) * (p^2 + q^2) - {q^3}{p^4} is odd.
  3.  p^3 + q^3 + 27p - 45q is even.

A. I only
B. II only
C. III only
D. I and II both
E. I and III both

Solution:

Step 1: Understand Question Statement
  • p and q are Prime numbers which are consecutive.
  • p < q

Of the given statement, we need to find out which must be true.

Step 2: Define Methodology
  • Since p and q are Primes which are consecutive and p < q, we can infer:
    • p = 2(i.e. p is even) and
    • q = 3 (i.e. q is odd)
  • Using the even-odd nature of p and q, let’s check which statement is always true
    • We will also use the fact that {(odd)}^n = odd and (even)^m = even, where n and m are any positive integer.
Step 3: Calculate the final answer

 I. {(7p + 9q)}^3 + 5q^2 - 36pq is even.

  • {(7p + 9q)}^3 + 5q^2 - 36pq = (odd*even + odd*odd) + odd*odd – even*even*odd ⟹ (even + odd) + odd – even = even

Hence, statement I must be true.

 II. {(105p + 280q)} * (p^2 + q^2) - q^3p^4 is odd.

  •   {(105p + 280q)} * (p^2 + q^2) - q^3p^4 = (odd*even + even*odd) *(even +odd) – odd*even ⟹ (even + even) *odd – even = even -even = even

Hence, statement II is not true.

 III. p^3 + q^3 + 27p - 45q is even.

  • p^3 + q^3 + 27p - 45q = even + odd + odd*even – odd*odd = odd +even – odd = even.

Hence, statement III must be true.

Thus, the correct answer is Option E.

Key learning from the solution

  • While solving the above question, students might skip the fact that 2 and 3 are only Prime numbers which are consecutive and mistake it as consecutive Prime numbers. Thus, they end up testing two cases:
    • Case 1: If 2 and 3 are consecutive Prime numbers. In this case, the result will be the same.
    • Case 2: If consecutive Prime numbers are other than 2 and 3. In this case, they may consider (3,5) or (5,7) and so on and notice both Primes are odd. So, the results of the expressions may be different.

Hence, on combining both the cases they might end up marking an incorrect answer.

So, always remember that 2 and 3 are only Prime numbers which are consecutive integers too.

Let’s solve another problem –

Example 2: (Scenario 2: In DS Question)

If Z = {P_1}^3 * {P_2}^2 * {P_3}^6 , is 7 a factor of Z?

  1. P_1, P_2, and P_3 are consecutive prime numbers.
  2. {P_1} and {P_2} are primes which are consecutive integers.

Solution –

Step 1: Analyse Question Stem
  • Z = {P_1}^3*{P_2}^2*{P_3}^6
  • We need to find if 7 is a factor of Z.
    • Or, at least one of {P_1}, {P_2}, and {P_3} is 7 or multiple of 7.
Step 2: Analyse Statements Independently (And eliminate options) – AD/BCE

Statement 1: {P_1}, {P_2}, and {P_3} are consecutive prime numbers.

  • According to this statement, there are multiple possible values for {P_1}, {P_2}, and {P_3} . Two such cases are given below:
    • Case 1:  {P_1}, {P_2}, and {P_3} can be 2, 3 and 5. In this case, Z won’t be divisible by 7.
    • Case 2:   {P_1}, {P_2}, and{P_3} can be 3, 5 and 7. In this case, Z will be divisible by 7.

As the above two results are contradictory, hence, statement 1 is not sufficient and we can eliminate answer options A and D

Statement 2:  P_1 and P_2  are Primes which are consecutive integers.

  • Since P_1 and P_2  are consecutive integers as well as Primes,
    • Hence, P_1 and P_2  must be 2 and 3.

However, we do not know about P_3  which may or may not be 7.

Hence, statement 2 is also not sufficient and we can eliminate answer B.

Step 3: Analyse Statements by combining.
  • From statement 1: P_1, P_2, and P_3  are consecutive Prime numbers.
  • From statement 2:  P_1 and P_2  are 2 and 3.
  • On combining both, we get : P_1, P_2, and P_3 are 2, 3 and 5.
    • So, we can infer that 7 is not a factor of Z.

Thus, the correct answer is Option C.

Key learning from the solution

  • In this question, knowing the property that 2 and 3 are only Primes which are consecutive integers helped.  Forgetting this may lead students to think that statement 2 is nothing but statement 1 written differently and they will end marking answer E.

Therefore, never confuse “consecutive Primes”  with “Primes which are consecutive integer” as the two phrases do not mean the same.

Prime Number’s Representation: (6n + 1) and (6n -1)- when and how to use

This is the final section of this article, in which we will see how to represent a Prime number and carefully understand when and how to use this representation.

Except for 2 and 3, all the Prime numbers can be represented in the form of either (6n+1) or (6n-1), where n is a positive integer.

  • For example: 6*3 +1=19 is a prime; 6*4 -1= 23 is also a Prime number.

Now, read the next statement very carefully:

Although all Prime numbers except 2 and 3 can be represented in these forms, not every number in the form of 6n+1 and 6n -1 is a prime.

  • For example:  6*8 +1 =49 is not a Prime; 6*11 -1= 65 is also not a Prime number.

Consequently, you must be cautious while using these forms. Now, the question comes, where to use and how?

So, keep the following in mind:

  • If you know you are dealing with a “Prime number greater than 3”, then you can represent that number as 6n + 1 or 6n – 1.
  • If you aren’t sure about the Prime nature of a number and it can be represented as 6n + 1 and 6n – 1, then it may or may not be a Prime number.
  • If you are dealing with Prime numbers in general, then make sure to consider 2 and 3 separately and then assume the other Primes as 6n+ 1 and 6n – 1.

Now, let’s take up two examples, to understand the application of this concept.

Example 1: (Scenario 1: Primes greater than 3)

If x is a Prime number greater than 3, what is the remainder when x^2 +723x  is divided by 3?

A. 1
B. 2
C. 3
D. 4
E. 5

Solution –

Step 1: Understand Question Statement
  • x is a Prime number.
  • x > 3
  • We need to find the remainder when x^2 + 723x is divided by 3.
Step 2: Define Methodology
  • One way to solve this problem is by plugging in values.
  • We can select any Prime number greater than 3 and can check for the remainder when x^2 + 723x  is divided by 3.

However, we will solve this methodically to understand the above property.

  • Look at the second term of the x^2 + 723x < i.e. 723x
    •  (7+2+3 = 12) which is divisible by 3 hence 723 is divisible by 3.
      • This means that the remainder of \frac {723x} {3} is 0.

So we only need to focus on the remainder of \frac {x^2} {3} and that will be the answer.

  • Since x > 3 and it is a Prime number.  So, we can represent x as:
    • x = 6n + 1 ⟹ {x}^2 = {(6n + 1)}^2= 36{n}^2  + 12n + 1 ……. Eq.(i)
    • Or, x = 6n – 1 ⟹ {x}^2 = {(6n - 1)}^2 = 36{n}^2 - 12n + 1 ……Eq.(ii)
    • Here n is a positive integer

Now, we will divide Eq.(i) and Eq.(ii) by 3 and see what comes as remainder.

Step 3: Calculate the final answer
  • Dividing Eq.(i) by 3
    •   \frac{x^2}{3} = \frac{36n^2 + 12n + 1}{3} = (12 * n^2 + 4 *n) + \frac{1}{3} . Therefore, the remainder is 1
  • Dividing Eq.(ii) by 12
    • \frac{x^2}{3} = \frac{36n^2 - 12n +1}{3} = (12 * n^2 - 4 * n) + \frac{1}{3} . Therefore, the remainder is 1

Results from both cases are the same.

Thus, the correct answer is Option A.

Key learning from the solution

  • In this problem, we were able to use (6n+1) and (6n-1) form to represent a Prime number because Prime numbers under consideration were greater than 3. If the same question would have been asked about all the Prime numbers, then we would not have been able to use these forms.

Example 2: (Scenario 2: where 6n+1 does not mean Prime)

Is p Prime?

  1.  p can be written in the form of 6n+1 where n is a positive integer.
  2. p^2  has exactly one factor between 1 and p^2

Solution –

Step 1: Analyse Question Stem
  • We need to find if p is a Prime.
Step 2: Analyse Statements Independently (And eliminate options) – AD/BCE

Statement 1:  p can be written in the form of 6n+1 where n is a positive integer.

  • We know that all Prime numbers except 2 and 3 can be represented in the form of 6n+1 and 6n-1(for example 13 = 6*2 +1 and 19 = 6*3+1), however, vice-versa need not to be true.
    • i.e. every number which can be represented in the form of 6n+1, where n is an integer, need no to be a Prime. For, example:
      • 25 = 6*4 +1, however 25 is not a Prime
      • 49 = 6*8 +1, this is also not a Prime.
  • Hence, statement 1 is not sufficient and we can eliminate answer options A and D

Statement 2:  p^2 has exactly one factor between 1 and p^2

  • According to this statement, p^2 has 3 factors.
    • Or, the total number of factors of p^2 = (2 + 1) = 3, this is possible only if p is Prime.
  • Hence, p is a Prime.

Thus, the correct answer is Option B.

Key learning from the solution

  • In this example, you noticed that the representation of a number in 6n+1 (or 6n -1) does not guarantee the prime nature of that number. If students have missed this concept, they will mark Option D as the correct answer, unfortunately, which is incorrect.

So, in this article, we learned 3 special properties of Prime numbers and their application. To avoid making mistakes in GMAT Quant, it is important to remember and apply these properties whenever you are solving the questions based on Prime numbers.

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