Numbers - Aptitude Questions and Answers

Numbers are the fundamental units of mathematics. In this article, we will discuss numbers, their types, important facts, divisibility rules, and other important points about numbers for aptitude preparation.

Types of Numbers

1. Integers: All numbers whose fractional part is 0 (zero) like -3, -2, 1, 0, 10, and 100 are integers.
2. Natural Numbers: Counting numbers like 1, 2, 3, 4, 5, 6 ... Basically, all integers greater than 0 are natural numbers.
3. Whole Numbers: All natural numbers and 0 (zero) are whole numbers.
4. Prime Numbers: All numbers having only two distinct factors, the number itself and 1, are called prime numbers. Some prime numbers are 2, 3, 53, 67 and 191.
5. Composite Number: All numbers greater than 1 which are NOT prime are composite numbers. Some composite numbers are 4, 60, 91 and 100.

Important Points on Prime Numbers

• 0 and 1 are neither prime nor composite.
• 2 is the only even prime number.
• There are 25 prime numbers less than 100. They are : 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97.
• To check if a number 'p' is prime, find a number 'n' such that 'n' is the smallest natural number that satisfies n² >= p. Now, check if 'p' is divisible by any of the prime numbers less than or equal to 'n'. If 'p' is NOT divisible by any such prime numbers, 'p' is a prime number. Otherwise, p is not a prime number.
• Co-primes : Two numbers 'a' and 'b' are called co-prime if their highest common factor (HCF) is 1.
• We can find the count of divisors of a number using prime factorization. If prime factorization of a number n is p₁^e1 x p₂^e2 x p₃^e3 ... p_k^ek. then the number of divisors of n are (e1 + 1) x (e2 + 1) x (e3 + 1) x .... (ek + 1). For example, 200 can be written as 2³5². The number of divisors of 200 is (3 + 1) x (2 + 1) = 12.

Read More: Interesting Facts on Prime Numbers and Prime Factorization Tips and Tricks for more details.

Divisibility Rules

Here's a table summarizing the divisibility rule of common numbers:

Other Useful Rules

• If we repeat a three-digit number twice, to form a six-digit number. The result will be divisible by 7, 11, and 13, and dividing by all three will give your original three-digit number.
• A number of form 2^N has exactly N+1 divisors.
  • For example: 4 has 3 divisors, 1, 2, and 4.

Read More: Divisibility Rules 1 to 19

Note : If 'p' and 'q' are co-primes and we have a number 'n' that is divisible by both 'p' and 'q', 'n' will be divisible by p x q. 

For example, 48 is divisible by both 3 and 8 and also by 3 x 8 = 24. 

But, if 'p' and 'q' are NOT co-prime, then the fact that 'n' would be divisible by p x q given that 'n' is divisible by both 'p' and 'q' is not necessary. For example, 144 is divisible by both 8 and 12 (not co-prime), but it is not divisible by 8 x 12 = 96. 

Remainder Facts

The table summarizes the remainder rules for different divisors, providing a quick and easy way to determine the remainder without performing full division. Each rule is accompanied by an example for clarity.

Division Theorem

Dividend = (Divisor x Quotient) + Remainder

1. (xⁿ - aⁿ) is divisible by (x - a) for all values of n. 

For example:

• For n = 2, x² - a² = (x - a) (x + a), which is divisible by (x - a). 
• Similarly, for n = 3, x³ - a³ = (x - a) (x² + a² + xa), which is divisible by (x - a).

2. (xⁿ - aⁿ) is divisible by (x + a) for all even values of n. 

For example:

• For n = 2, x² - a² = (x - a) (x + a), which is divisible by (x + a). 
• Similarly, for n = 3, x³ - a³ = (x - a) (x² + a² + xa), which is not divisible by (x + a).

3. (xⁿ + aⁿ) is divisible by (x + a) for all odd values of n. 

For example:

For n = 3, x³ + a³ = (x + a) (x² + a² - xa), which is divisible by (x + a).

Check: Euclid Division Lemma

Cyclicity of Numbers

The cyclicity of any number is mainly focused on its unit digit. Every unit digit has its own repetitive pattern when raised to any power.

• Digits 0, 1, 5, and 6: Here, when each of these digits is raised to any power, the unit digit of the final answer is the number itself.

Examples:

1. 5² = 25: Unit digit is 5
2. 1⁶ = 1: Unit digit is 1
3. 0⁴ = 0: Unit digit is 0
4. 6³ = 216: Unit digit is 6

• Digits 4 and 9: Both of these two digits, 4 and 9, have a cyclicity of two different digits as their unit digit.

Examples:

4² = 16: Unit digit is 6.
4³ = 64: Unit digit is 4.
9² = 81: Unit digit is 1.
9³ = 729: Unit digit is 9.

• Digit 2 has a cyclicity of four different numbers: 2,4,8,6.
• Digit 3 has a cyclicity of four different numbers: 3, 9, 7, 1.
• Digit 7 has a cyclicity of four different numbers: 7, 9, 3, 1.
• Digit 8 has a cyclicity of four different numbers: 8, 4, 2, 6.

Read in Detail: Cyclicity of Numbers

Sample Problems on Number System

Question 1: When a number is successively divided by 2, 3, 7 we get 1, 2, and 3 as the remainder respectively. What is the smallest such number?
Solution:

The number is of the form 2a + 1, 3b + 2, 7c + 3. So, we put c = 1 and proceed as follows :

Basically, we successively multiply the divisors with the result of the previous stage and add the corresponding remainder. 
7 x 1 + 3 = 10 
3 x 10 + 2 = 32 
2 x 32 + 1 = 65 
Thus, 65 is the required answer. 
NOTE : The answer would differ if we change the order of divisors. For smallest number, arrange the divisors in decreasing order. 

Question 2: When a number is successively divided by 2, 4, 8 we get 1, 1, and 0 as the remainder respectively. What is the smallest such number? 
Solution:

Proceeding in the similar manner as the above question, 
8 x 1 + 0 = 8 
4 x 8 + 1 = 33 
2 x 33 + 1 = 67 
Thus, 67 is the required answer. 

Question 3: What would be the maximum value of 'B' in the following equation : 

1 2 B
+ B 4 C
+ C 6 7
--------
1035
--------
Solution:

Only the leftmost part of the number can be of two or more digits. So, we split the answer as : 

1 2 B
+ B 4 C
+ C 6 7
--------
10 3 5
--------

Now, from column 1, we can easily infer that B + C = 8. 
First, let us consider B + C = 18. This is the case possible if and only if B = C = 9. So, the equation would be 129 + 949 + 967 = 2045, but we need 1035 as the answer. Thus, this is not the required case. 
So, B + C = 8. For maximum 'B', we put C = 0. Therefore, B = 8. 
Now, to verify our answer, we put B = 8 and C = 0 in the given equation. 

1 2 8
+ 8 4 0
+ 0 6 7
--------
10 3 5
--------

Therefore, our answer B = 8 is correct. 

Question 4: Which of the following are prime numbers? 
(i) 247 
(ii) 397 
(iii) 423 

Solution:

(i) 16² = 256 > 247. Prime numbers less than 16 are 2, 3, 5, 7, 11, 13 and 247 is divisible by 13. Therefore, 247 is not a prime number. It is a composite number. 
(ii) 20² = 400 > 397. Prime numbers less than 20 are 2, 3, 5, 7, 11, 13, 17, 19 but 397 is not divisible by any of these. Therefore, 397 is a prime number. 
(iii) 21² = 441 > 423. Prime numbers less than 21 are 2, 3, 5, 7, 11, 13, 17, 19 and 423 is divisible by 3. Therefore, 423 is not a prime number. It is a composite number. 

Question 5: Find the unit's digit in the product (17)¹⁵³ x (31)⁶². 
Solution:

The unit's digit of the given equation would be the same as the unit's digit of the equation 7¹⁵³ x 1⁶². 
Now, we need to find a pattern in the unit's digit when we gradually increase the powers of 7. 7¹ gives 7, 7² gives 9, 7³ gives 3, 7⁴ gives 1. So, at the fourth power, we get the unit's digit as 1. Therefore, to make our work easy, we need to write the original power (153) in multiples of 4 to the closest extent possible. We multiply this power (4) by a number such that we get nearest to 153. So, 4 x 38 = 152 and 7¹⁵² also has 1 in the unit's place. 
Now, (17)¹⁵³ has 7 at unit's place and (31)⁶² has 1 at the unit's place. 
Therefore, the problem simply reduces to 7 x 1 = 7. 
Hence, the unit's digit is 7. 

Question 6: Find the unit's digit in (17)¹⁵³ + (31)⁶². 
Solution:

The unit's digit of the given equation would be the same as the unit's digit of the equation 7¹⁵³ + 1⁶². 
Now, we need to find a pattern in the unit's digit when we gradually increase the powers of 7. 7¹ gives 7, 7² gives 9, 7³ gives 3, 7⁴ gives 1. So, at the fourth power, we get the unit's digit as 1. Therefore, to make our work easy, we need to write the original power (153) in multiples of 4 to the closest extent possible. We multiply this power (4) by a number such that we get nearest to 153. So, 4 x 38 = 152 and 7¹⁵² also has 1 in the unit's place. 
Now, (17)¹⁵³ has 7 at unit's place and (31)⁶² has 1 at the unit's place. 
Therefore, the problem simply reduces to 7 + 1 = 8. 
Hence, the unit's digit is 8. 

Question 7: Find the total number of prime factors in the expression (14)¹¹ x (7)² x (11)³. 
Solution:

(14)¹¹ x (7)² x (11)³ = (2 x 7)¹¹ x (7)² x (11)³ = (2)¹¹ x (7)¹¹ x (7)² x (11)³ = (2)¹¹ x (7)¹³ x (11)³ 
Therefore, total number of prime factors = 11 + 13 + 3 = 27 

Question 8: Which digits should come in place of * and # such that the number 12386*# is divisible by both 8 and ? 
Solution:

Since the given number should be divisible by 5, 0 or 5 must come in place of #. But, a number ending with 5 is never divisible by 8. So, 0 will replace #. 
Now, the number formed by the last three digits is 6*0, which becomes divisible by 8, if * is replaced by 0 or 4 or 8. 
Hence, digits in place of * and # are 0 or 4 or 8 and 0 respectively. 

Question 9: What is the least number that must be subtracted from 9999 to make it exactly divisible by 19? 
Solution:

On dividing 9999 by 19, we get 5 as remainder. Therefore, number to be subtracted = 5. 

Question 10: What is the least number that must be added to 9999 to make it exactly divisible by 19 ?
Solution:

On dividing 9999 by 19, we get 5 as remainder. Therefore, number to be added = 19 - 5 = 14. 

Question 11: A number when divided by 340 gives a remainder of 47. What would be the remainder when the same number is divided by 17? 
Solution:

The number is of the form 340a + 47 = 17 * (20a) + 17 * (2) + 13 = 17 * (20a + 2) + 13. 
Therefore, on dividing this number by 17, we would get 13 as the remainder. 

Question 12: Find the remainder when 3²¹ is divided by 5. 
Solution:

3⁴ = 81. So, the unit's digit of 3⁴ is 1. 
Therefore, the unit's digit of 3²⁰ = 1 and thus, the unit's digit of 3²¹ = 1*3 = 3. 
3 when divided by 5 gives 3 as the remainder. 
So, the remainder when 3²¹ is divided by 5 is 3. 

Also Check:

• Aptitude Questions and Answers
• Quiz about Numbers
• Quantitative Aptitude - Topics, Questions and Answers
• Number Series Reasoning Questions and Answers
• Cyclicity of Numbers