Mathematics · Quantitative Aptitude

Number Theory

466 Questions

Number theory explores the properties and relationships of integers, focusing heavily on prime and composite numbers. Key areas include identifying prime numbers, understanding co primes, and exploring mathematical conjectures like Goldbach's. These questions are a core component of quantitative aptitude sections in banking and government exams.

Prime number identificationCo prime propertiesGoldbach conjectureOdd and even numbersDivisibility rules

Number Theory Questions

Multiple choice mathematics and statistics introduction to set union and intersections union and intersection of sets basic operations on sets

Let A = { even number} B = {prime numbers} Then A $\displaystyle\cap $ B equals: 

  1. {odd number}

  2. {composite number}

  3. {2}

  4. {whole numbers}
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given: A = ${2, 4, 6, ...}$
     B = ${2, 3, 5, ...}$
$\displaystyle \therefore $ 2 is the only even prime number $\displaystyle A\cap B=\left { 2 \right }$

Multiple choice maths equations and simple functions further equations solving linear equations with variable on both sides solution of linear equations in one variable

The sum of three non-zero prime numbers is $100$. One of them exceeds the other by $36$. Find the largest number.

  1. $73$
  2. $91$
  3. $67$
  4. $57$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
As we know that the sum of three odd numbers cannot be even.
$\therefore$ one of the prime is even 
Since $2$ is the only prime number which is even.
$\therefore$ one of the three prime numbers is $2$.
Let one of the other prime numbers is ${p} _{1}$ then the third prime number will be ${p} _{1} + 36$. 
Now according to question,
$2 + {p} _{1} + {p} _{1} + 36 = 100$
$2{p} _{1}=62$
$p1=31$

Hence, the three prime numbers are $2, 31$ and $67$ and the largest among them is $67$.
Hence, $67$ is the correct answer.

Multiple choice

Which number is prime?

  1. 14

  2. 12

  3. 13

  4. 15
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

A prime number is a natural number greater than 1 that has no positive divisors other than 1 and itself. 13 is only divisible by 1 and 13.

Multiple choice

Which number is prime?

  1. 32

  2. 2

  3. 49

  4. 1230
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

A prime number is a natural number greater than 1 that has no positive divisors other than 1 and itself. 2 is the only even prime number.

Multiple choice maths rational numbers proof of irrationality of numbers proofs of irrationality introduction to irrational numbers

Which of the following is always true 

  1. $irrational + irrational =irrational $
  2. $\dfrac{rational }{rational }=rational $
  3. $\dfrac{integer }{integer}=integer$

  4. None of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Counter-example for A: $(\sqrt{2}) + (4-\sqrt{2}) = 4$

Counter-example for C: $\dfrac{1}{2}=0.5$

Proof for B:

Let $q _1, q _2$ be two rational numbers such that $q _2\neq0$.


As they are rational, they can be written as $a/b, c/d$ respectively for some integers $a, b, c, d$. $(b,c,d\neq0)$

$\dfrac{q _1}{q _2}=\dfrac{a/b}{c/d}=\dfrac{ad}{bc}$

Since, $a,b,c,d$ were integers, even $ad$ and $bc(\neq0)$ are integers and therefore, the above expression is rational.

Multiple choice maths rational numbers proof of irrationality of numbers proofs of irrationality introduction to irrational numbers

Every irrational number is

  1. a surd

  2. a prime number

  3. not a surd

  4. none
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

An irrational number is a real number that cannot be represented as a ratio or a simple fraction.


By definition, a surd is an irrational root of a rational number. So we know that surds are always irrational and they are always roots.

For eg, $\sqrt2$ is a surd since 2 is rational and $\sqrt 2$ is irrational.

Similarly, the cube root of 9 is also a surd since 9 is rational and the cube root of 9 is irrational.

On the other hand, $\sqrtπ$ is not a surd even though $\sqrtπ$ is irrational because π is not rational.

Thus, to answer the question, every surd is an irrational number, though an irrational number may or may not be a surd


The answer is Option C.

Multiple choice finding nth roots of a complex number n th root of unity demoivre's theorem complex numbers maths

If $p$ and $q$ are distinct prime numbers, then the number of distinct imaginary numbers which are $p$th as well as $q$th roots of unity are

  1. min$(p, q)$
  2. max$(p, q)$
  3. $1$

  4. zero
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

It is given that, $p$ and $q$ are prime numbers.
Hence the only common $pth$ and $qth$ root of unity will be the number 1.
Thus there will be no common imaginary $pth$ and $qth$ root of unity.
Hence answer is zero.

Multiple choice maths decimal numbers adding and subtracting decimals addition and subtraction of decimals operations on decimals

A mathematician born in the first half of the 19th century was x years old in the year $x^2$. He was born in

  1. 1849

  2. 1806

  3. 1812

  4. 1852
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The man born between $1800$ and $1850$ which is a perfect square. 
The perfect square number is $43 = 1849$ (Since$ 42 = 1764$ and $44 = 1936$ ) 
So in the year $1849$ the man was $43$ years old. 
Which shows the year of born is $1849 - 43 = 1806.$

Multiple choice maths rational and irrational numbers existence of irrational numbers irrational numbers properties of irrational numbers

If $p$ is prime, then $\sqrt {p}$ is:

  1. Composite number

  2. Rational number

  3. Positive integer

  4. Irrational number
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

SInce, we know that prime numbers are those which are never perfect square and not divisible by any other number except by itself.
which are $2,3,5,7,...$
Clearly, if $p$ is prime then $\sqrt p $ is irrational number.
Option $D$ is correct. 

Multiple choice

The successor of every even number is

  1. Even

  2. Prime

  3. Odd

  4. Composite
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Even numbers follow the pattern 2n. The successor is 2n + 1, which is the definition of an odd number.