Number Theory Questions

Number theory questions and answers are available on this page and contain various types of questions. These questions cover different types of techniques and formulas in mathematics and are particularly related to numbers. All these number theory questions will help you to understand how to solve complex problems using simple shortcuts and tricks.

What is the number theory?

Number theory is one of the elementary branches of mathematics that deals with the study of numbers (natural numbers) and properties of numbers, classification of numbers based on certain arithmetic operations.

Also read: Number theory

Number Theory Questions and Answers

1. Find the number of trailing zeros in the 100!.

Solution:

Let us divide 100 by 5.

100/5 = 20

20/5 = 4

Now, 4 is less than 5. So, we stop the division here.

Also, the number of trailing zeros = 20 + 4 = 24

2. Why is 1 not considered to be prime?

Solution:

From the definition of prime numbers, we can say that a number is considered to be prime if it contains two distinct factors, namely, 1 and the number itself. The number 1 is divided by 1 itself. Thus, it cannot have any other factor. Hence, 1 is not considered to be prime.

3. Find the LCM of 8, 27 and 72 using the prime factorisation method.

Solution:

Prime factorisation of 8 = 2 × 2 × 2

Prime factorisation of 27 = 3 × 3 × 3

Prime factorisation of 72 = 2 × 2 × 2 × 3 × 3

LCM(8, 27, 72) = 2 × 2 × 2 × 3 × 3 × 3 = 216

Therefore, the LCM of 8, 27 and 72 is 216.

4. Write the first 10 non-zero multiples of 7.

Solution:

Multiple of a number is the result obtained when the number is multiplied by integers.

The first 10 non-zero multiples of 7 are written as follows:

7 × 1 = 7

7 × 2 = 14

7 × 3 = 21

7 × 4 = 28

7 × 5 = 35

7 × 6 = 42

7 × 7 = 49

7 × 8 = 56

7 × 9 = 63

7 × 10 = 70

Thus, the first 10 non-zero multiples of 7 are 7, 14, 21, 28, 35, 42, 49, 56, 63 and 70.

5. What is the value of (61² − 39²) ÷ (51² − 49²)?

Solution:

(61² − 39²) ÷ (512 − 492)

Consider (61² – 39²)

Using the identity a² – b² = (a + b)(a – b)

(61² – 39²) = (61 + 39) (61 – 39)

= 100 × 22

(51² – 49²) = (51 + 49) (51 – 49)

= 100 × 2

Therefore, (61² − 39²) ÷ (51² − 49²) = (100 × 22)/ (100 × 2) = 11

6. What are the common factors of 48 and 54?

Solution:

Factors of 48 = 1, 2, 3, 4, 6, 8, 12, 16, 24, 48

Factors of 54 = 1, 2, 3, 6, 9, 18, 27, 54

Here, 1, 2, 3, and 6 are common for both 48 and 54.

Thus, the common factors of 48 and 54 are 1, 2, 3 and 6.

7. Write the first three common multiples of 15 and 20.

Solution:

The multiples of 15 are 15, 30, 45, 60, 75, 90, 105, 120, 135, 150, etc.

The multiples of 20 are 20, 40, 60, 80, 100, 120, 140, 160, 180, 200, etc.

The first three common multiples of 15 and 20 are 60, 90 and 120.

8. Prove that the sequence 2, 5, 8, 11, 14, 17,.… can never have a square number.

Solution:

Given,

2, 5, 8, 11, 14, 17,.…

This is an AP with the first term a = 2 and common difference d = 5 – 2 = 3.

nth term of the given sequence is:

a_n = a + (n – 1) × d

a_n = 5 + (n – 1) × 3

= 5 + 3n – 3

= 3n + 2

∴ a_n = 3n + 2

Let p be a natural number, such that p² = an

p² = 3n + 2

3n = p² – 2

n = (p² – 2)/3

For any integer from 0 to 9 for p, n does not appear to be an integer.

Hence, the given sequence contains no perfect squares.

9. Find all primes that can be represented as sums and differences between two primes.

Solution:

Let x be a prime that can be represented both as a sum and as a difference of 2 primes.

For the given statement, we must have x > 2.

∴ x is an odd prime number.

Also, one of those prime numbers must be 2.

That means we must have x = y + 2 = z – 2, where y and z are prime numbers.

So, z = x + 2 and y = z + 2

Thus, we can say that x, y and z will be three consecutive odd primes.

As we know, there is only one such set of prime numbers, and they are 3, 5, and 7.

x = 5 = 3 + 2 = 7 – 2

Hence, there is only one prime number .i.e 5 can be represented as sums and differences between two primes.

10. What is the divisibility rule of 11? Give one example.

Solution:

If the difference between the sum of digits of a number at odd places and the sum of digits of the number at even places is equal to 0 or a multiple of 11, then the number is divisible by 11.

Example:

Consider the number 719752

Sum of digits at odd places = 7 + 9 + 5 = 21

Sum of digits at even places = 1 + 7 + 2 = 10

Difference = 21 – 10 = 11 (multiple of 11)

Therefore, the number 719752 is divisible by 11.

Practice Questions on Number Theory

1. If the product of the integers w, x, y and z is 770, and if 1 < w < x < y < z, what is the value of w + z?
2. Find the GCF of 67 and 194 using the prime factorisation method.
3. Identify the number of zeros in the value of 1000!.
4. What is the divisibility rule for 7?
5. What is the largest five-digit number that is divisible by 7, 10, 15, 21 and 28?