Quantitative Aptitude · Mathematics

Numbers and Divisibility

209 Questions

Master the number system by solving these quantitative aptitude questions on divisibility rules and properties. The exercises cover finding the greatest common divisor and identifying prime factors. This topic is essential for clearing the preliminary stages of SSC, banking, and various state exams.

Divisibility rulesGreatest common divisorPolynomial divisionNatural numbersFactorizationPrime numbers

Numbers and Divisibility Questions

Multiple choice maths numbers and place value forming numbers formation of greatest and smallest numbers identifying the largest and smallest numbers with given digits

A three digit number from the given digits $2, 5, 7,9$ which divisible by 2.

  1. $257$
  2. $925$
  3. $527$
  4. $752$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Consider the given digits.

$2, 5, 7, 9$

If a number which is divided by $2$, then the units digit must be even.

So, from the given options there is only one option which has unit digit $2$.

Hence, the three digit number will be $752$.

Hence, this is the answer.

Multiple choice maths numbers and place value forming numbers formation of greatest and smallest numbers identifying the largest and smallest numbers with given digits

Find a three-digit number using the digits $6, 7, 4$ such that the resultant number is divisible by 4

  1. $746$
  2. $764$
  3. $467$
  4. $647$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
(a) Let us take a number $746$ using the digits $6,7,4$ and divide it by $4$ as follows:

$\dfrac {746}{4}=\dfrac {373}{2}$

Therefore, $746$ is not divisible by $4$.

(b) Let us take a number $764$ using the digits $6,7,4$ and divide it by $4$ as follows:

$\dfrac {764}{4}=191$

Therefore, $764$ is divisible by $4$.

(c) Let us take a number $467$ using the digits $6,7,4$ and divide it by $4$ as follows:

$\dfrac {467}{4}$

Therefore, $467$ is not divisible by $4$.


(d) Let us take a number $647$ using the digits $6,7,4$ and divide it by $4$ as follows:

$\dfrac {647}{4}$

Therefore, $647$ is not divisible by $4$.

Hence, only $764$ is divisible by $4$.
Multiple choice lines in planes applications of determinants inverse of a matrix and linear equations matrix algebra maths

Three digits numbers $ 7x,36y$  and  $12z$ where  $x , y , z$ are integers from  $0$  to  $9 ,$  are divisible by a fixed constant $k.$  Then the determinant  $\left| \begin{array} { l l l } { x } & { 3 } & { 1 } \ { 7 } & { 6 } & { z } \ { 1 } & { y } & { 2 } \end{array} \right|$ $\ +48$ must be divisible by 

  1. $k$
  2. $k ^ { 2 }$
  3. $k ^ { 3 }$
  4. $k ^ { 4}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Since $7x,36y,12z$ are divisible by $k$

Let us Assume $x=ka,y=kb,z=kc$
So the det $\implies \left| \begin {array}{c c c} ka&3&1\7&6&kc\1&kb&2 \end{array} \right|$
$\implies 12ka-k^3abc+3kc+7kb-48$
$\implies From \  Question, 12ka-k^3abc+3kc+7kb-48+48$
Hence A

Multiple choice lines in planes applications of determinants inverse of a matrix and linear equations matrix algebra maths

Numbers of ways in which 75600 can be resolved as product of two divisors which are relatively prime ?

  1. 44
  2. $8$
  3. $9$
  4. $16$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

First we will have to find the prime factors of $75600$

Prime factorization of $75600=2\times 2\times 2\times 2\times 3\times 3\times 3\times 5\times 5\times 7$
$\Rightarrow 75600=2^4\times 3^3 \times 5^2\times 7$
The number of ways in which a composite number N can be resolved as product of two divisors which are relatively prime.
$=2^{n-1}$ where n is number of different factors of N
$=2^{n-1}$
$=2^{4-1}$
$=2^3$
$=8$ ways
Hence, the answer is $8.$

Multiple choice statistics information processing fundamental principle of addition fundamental principles of counting principles of counting

Total number of ways of selecting two numbers from the set ${1,2,3,...90}$ so that their sum is divisible by $3$ is

  1. $885$
  2. $1335$
  3. $1770$
  4. $3670$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Numbers 1 to 90 can be classified by remainder when divided by 3: 30 numbers each in categories congruent to 0, 1, and 2 mod 3. For sum of two numbers to be divisible by 3: either both are ≡ 0 (mod 3), or one is ≡ 1 and other ≡ 2 (mod 3). Ways from first category: C(30,2) = 30×29/2 = 435. Ways from second category: 30 × 30 = 900. Total ways = 435 + 900 = 1335.

Multiple choice maths real number fundemental theorem of arithmetic real numbers on number line fundamental theorem of arithmetic

We know that any odd positive integer is of the form $4q + 1 $ or $4q + 3$ for some integer $q.$
Thus, we have the following two cases.

  1. $n^2-1$ is divisible by 8
  2. $n^2+1$ is divisible by 8
  3. $n-1$ is divisible by 8
  4. $n+1$ is divisible by 8
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

When $n=4q+1$
In this case, we have
$n^2-1=(4q+1)^2-1=16q^2+8q+1-1$
$=8q(2q+1)=8r$ where $r=q(2q+1)$ is an integer
$\Rightarrow n^2-1$ is divisible by 8.
Case-II: When $n=4q+3$
In this case, we have
$n^2-1=(4q+3)^2-1=16q^2+24q+9-1=16q2+24q+8$
$=8(2q^2+3q+1)=8(2q+1)(q+1)$
$=8r$ where $r=(2q+1)(q+1)$ is an integer.
$\Rightarrow n^2-1$ is divisible by 8
Hence $n^2-1$ is divisible by 8.

Multiple choice maths real number fundemental theorem of arithmetic real numbers on number line fundamental theorem of arithmetic

 One and only one out of  $n, n + 4, n + 8, n + 12\  and \ n + 16 $ is ......(where n is any positive integer)

  1. Divisible by 5

  2. Divisible by 4

  3. Divisible by 10

  4. Divisible by 12
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

We know that any positive integer is of the form 5q, 5q + 1 or 5q + 2, 5q + 3 or 5q + 4 for some integer q and one and onlyone of these possibilities can occur. So, we have the following cases:
Case-I: When $n=5q$
In this case, we have
$n=5q$, which is divisible by 5
Now, $n=5q$
$\Rightarrow n+4=5q+4$
$\Rightarrow n+4$ leaves remainder 4 when divided by 5
$\Rightarrow n+4$ is not divisible by 5.
Now $n+8=5q+8=5(q+1)+3=5m+3$, m is an integer.
Clearly, n+8 is not divisible by 5.
Again, $n+12=5q+12=5(q+2)+2=5m+2$, m in an integer.
Clearly n+12 is not divisible by 5.
Now $n+16=5q+16=5(q+13)+1=5m+1$, m is an integer
$\Rightarrow n+16$ is not divisible by 5
Thus, if n = 5q only one out of n, n + 4, n + 8, n +
12 and n + 16 is divlsible by 5,
Similarly, this result can be proved for the rest of .
the cases.

Multiple choice maths be my multiple, i'll be your factor co-prime numbers lcm lowest common multiple (l.c.m.)

The greatest number which when subtracted from 5834, gives a number exactly divisible by each of 20, 28, 32 and 35 is

  1. 1120

  2. 4714

  3. 5200

  4. 5600
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

 Number which is exactly divisible by 20, 28, 32 and 35 should be the common multiple of all these.
$20 = 2^2 \times 5$
$28 = 2^2 \times 7 \Rightarrow 32 = 2^5$
35.=5 $\times $ 7
LCM = $2^5 \times 5 \times 7 = 1120$
Hence the greatest number that should be subtracted
=5834-1120 = 4714

Multiple choice maths negative numbers and integers even and odd numbers sum of numbers odd and even numbers

Given that a, b are odd and c, d are even. Then,

  1. $\displaystyle a^{2}-b^{2}+c^{2}-d^{2}$ is always divisible by 4
  2. $abc + bcd + cda + dac$ is always divisible by 4
  3. $\displaystyle a^{4}+b^{4}+c^{3}+d^{3}+c^{2}b+a^{2}b$ is always odd
  4. $a + 2b + 3c + 4d$ is odd
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Let $a=1, b= 3, c= 2, d = 4$
Option D
$a+2b+3c+4d$
1+6+6+16 =29 which is odd number
In other 3 option always not correct for different values for a, b , c, d

Multiple choice maths negative numbers and integers even and odd numbers sum of numbers odd and even numbers

If $P$ is an integer between $0$ and $9,R-P=16229$ and $R$ divisible by $11$, then find the value of $\dfrac {P+R-1}{3}$

  1. $5014$
  2. $4514$
  3. $5414$
  4. $5114$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
$ R-P = 16229 $
P be b/w $0\& 9 $
and R is divisible by 11
So, $ R = \dfrac{16229+P}{11} $
(and Reminder = 0)
So, $ \Rightarrow (\dfrac{11+P}{11}) $ so $ P = 7 $
and $ R = 16229+7 $
$ = 16236 $
So $ \dfrac{P+R+1}{3} $
$ \Rightarrow \dfrac{16236+7-1}{3} $
$ \Rightarrow \dfrac{16242}{3} = 5414 $ 
Option C is correct