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 negative numbers and integers even and odd numbers sum of numbers odd and even numbers

Consider $n={21}^{52}$, then

  1. number of even divisors of $n$ is $704$
  2. number of odd divisors of $n$ is $2809$
  3. last two digits of $n$ is $41$
  4. number of even divisors of $n$ which are multiple of $9$ is $2705$
Reveal answer Fill a bubble to check yourself
B,C Correct answer
Explanation

Let,we have


$n = {21^{52}}$

can be written as $n = {\left( {7 \times 3} \right)^{52}}$

$n = {7^{52}}{.3^{52}}$

We know, no. of total divisors of any number

$k = {p^m}.{q^n}$

Total divisors$=(m+1)\,(n+1)$

so, similary here

odd divisors$=(52+1)(52+1)=2809$

Hence the option $(B)$ is correct

But again 

For last two digit

$n = {21^{52}} = {\left( {20 + 1} \right)^{52}}$

${\left( {20 + 1} \right)^{52}}{ = ^{52}}{C _1}{\left( {20} \right)^{52}} + .....{ + ^{52}}{C _{51}}{\left( {20} \right)^1}{ + ^{52}}{C _{52}}{\left( {20} \right)^0}$

For last two digit we notice last two terms 

$=^{52}{C _{51}}\left( {20} \right) + 1$

$ = 52 \times 20 \times 1$

$=1041$

$1041$ has last two digit is $41$

so, option $(C)$ is also correct

Hence both the option $(B)$ and $(C)$ are correct.

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

The integer just below $(\sqrt{53}+7)^{11}-2\times 7^{11}$ is 

  1. Divisible by exactly $4$ primes factors
  2. Divisible by exactly $3$ primes factors
  3. is divisible by $7$
  4. has $53$ as its only two digit prime factor
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The expression (sqrt(53)+7)^11 - 2*7^11 involves binomial expansion. The term (sqrt(53)+7)^11 can be written as (7+sqrt(53))^11 + (7-sqrt(53))^11, which is an integer. The value is approximately 2*7^11, and the integer just below it relates to the properties of these powers.

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

The number of even proper divisor of 1008 is

  1. 18

  2. 17

  3. 23

  4. 9
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

We have,

$ 1008=2\times 2\times 2\times 2\times 3\times 3\times 7 $

$ ={{2}^{4}}\times {{3}^{2}}\times {{7}^{1}} $

$ =2\left( {{2}^{3}}\times {{3}^{2}}\times {{7}^{1}} \right) $

Then, the number of even divisors

$ =\left( 3+1 \right)\left( 2+1 \right)\left( 1+1 \right) $

$ =24 $

But the above divisors also contain the 1008 which is not a proper divisors

Then number of proper divisors$=23$

Hence, this is the answer.

Multiple choice business maths probability - i addition theorems of probability statistics and probability descriptive statistics and probability

An integer is chosen at random from the first two hundred digits.What is the probability that the integer chosen is divisible by 6 or 8 ?

  1. $\displaystyle\frac{29}{100}.$
  2. $\displaystyle\frac{1}{4}.$
  3. $\displaystyle\frac{1}{8}.$
  4. $\displaystyle\frac{21}{100}.$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let A=the integer is divisible by 6 
$A={6,12,18,....198}$
$198=6+(n-1)6$
$\Rightarrow n =33$
So, $P(A)=\dfrac{33}{200}$

B=the integer is divisible by 8
$B={8,16,24,....200}$
$200=8+(n-1)8$
$\Rightarrow n=25$
So, $P(B)=\dfrac{25}{200}$

Both are divisible by 6 and 8 both 
$A\cap B=24,48,.....192$
$192=24+(n-1)24$
$\Rightarrow n=8$
So, $P(A\cap B)=\dfrac{8}{200}$

$\displaystyle \therefore P\left ( A\cup B \right )=P\left ( A \right )+P\left ( B \right )-P\left ( AB \right )$
$\displaystyle=\frac{33}{200}+\frac{25}{200}-\frac{8}{200}=\frac{1}{4}.$

Multiple choice maths introduction to euclid's geometry conditional statements and converse euclid's postulates axioms, postulates and theorems euclid's fifth postulate

Using Euclid's Division Lemma, for any positive integer $n, n^3-n$ is always divisible by 

  1. $6$
  2. $4$
  3. $3$
  4. $8$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$n^{3}-n=n(n^{2}-1)=n(n-1)(n+1)$ is divisible by $3$ then  possible remainder is $0, 1$ and $2$


[$\because$ if $P=ab+r$, Then $0\le r < a$ by Euclid lemma]


$\therefore$ Let $n=3r, 3r+1,3r+2$ where $r$ is an integer

Case $1$: When $n=3r$

Then, $n^{3}-n$ is divisible by $3$  [$\because n^{3}-n=n(n-1)(n+1)=3r(3r-1)(3r+1)$, early shown it is divisible by $3$]

Case $2$: When $n=3r+1$

$n-1=3r+1-1=3r$

Then, $n^{3}-n=(3r+1)(3r)(3r+2)$ it is divisible by $3$

Case: when $n=3r-1$

$m+1=3r-1+1=3r$

Then, $n^{3}-n=(3r-1)(3r-2)(3r)$ it is divisible by $3$

Now out of three $(n-1)^{n}$ and $(n+1)$ are must be even so it is divisible by $2$

$n^{3}-n$ is divisible by $2\times 3=6$

Multiple choice maths geometric sequences sum of terms of g.p sum of n terms of an gp summing geometric series

Find the sum of odd integers between $1$ and $1000$ which are divisible by $3$.

  1. $83667$
  2. $54954$
  3. $99994$
  4. $79894$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Odd integers divisible by 3 are: $3,9,15,21.....999$
Let $a=3$   and   $d=6$
$\therefore T _n=999=3+(n-1)6$
$\Rightarrow n-1=\dfrac{996}{6}=166$
$\therefore n=167$
$\therefore S _n=\dfrac{167}{2}[2\times 3+(167-1)6]=83667$
Multiple choice maths hcf-lcm common factors and hcf hcf highest common factor (h.c.f.)

The two numbers nearest to 10000 which are exactly divisible by each of 2, 3, 4, 5, 6 and 7, are _____.

  1. 9660, 10080

  2. 9320, 10080

  3. 9660, 10060

  4. 10340, 10080
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The numbers which are exactly divisible by 2, 3, 4, 5, 6 and 7 are the multiples of the LCM of the given numbers.
$\therefore$  LCM = 2 x 2 x 3 x 5 x 7 = 420
Now, dividing 10000 by 420, we get remainder = 340
$\therefore$  Number just less than 10000 and exactly divisible by the given numbers = 10000 - 340 = 9660
Number just greater than 10000 and exactly divisible by the given numbers = 10000 + (420 - 340) = 10080

Multiple choice maths hcf-lcm common factors and hcf hcf highest common factor (h.c.f.)

The greatest common divisor of $878787878787$ and $787878787878$ equals.

  1. $3$
  2. $9$
  3. $27$
  4. $101010101010$
  5. $303030303030$
Reveal answer Fill a bubble to check yourself
E Correct answer
Explanation

$787878787878)878787878787(1\ \quad \quad \quad \quad \quad  -\underline { 787878787878 } \ \quad \quad \quad \quad \quad \quad \quad 90909090909)787878787878(8\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad  \underline { -727272727272 } \ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad 60606060606)90909090909(1\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \underline { -60606060606 } \ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad 30303030303)60606060606(2\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \underline { -60606060606 } \ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad  \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad 0$

$\therefore$ G.D.C = 30303030303

Multiple choice maths proofs in mathematics implications principle of mathematical induction different forms of theoretical statements

What is the symbolic form and truth value of the following?
"If $4$ is an odd number, then $6$ is divisible by $3$." 
p: $4$ is an odd number.
q: $6$ is divisible by $3$.

  1. p$\rightarrow$q and $F$
  2. q$\rightarrow$p and $T$
  3. q$\rightarrow$p and $F$
  4. p$\rightarrow$q and $T$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$p: 4$ is an odd number.
$q: 6$ is divisible by $3$.
Symbolic form: $p$ $\rightarrow$ $q$
$p$ is false and $q$ is true.
So, $F\rightarrow T$ is $T$.

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

Number of integers lying between $1 $ to $102$  which are divisible by all $\displaystyle \sqrt{2},\sqrt{3},\sqrt{6}, $ is 

  1. $16$
  2. $17$
  3. $15$
  4. $0$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

For a number to be divisible by $\sqrt { 2 } $, it must be an irrational number. An integer is not an irrational,

so  there are no  numbers between  $ 1$ to  $102$ which are divisible by all  $\sqrt{2},\sqrt{3},\sqrt{6}$.