Tag: maths

Questions Related to maths

If $f(x)=x^{2}+6x+c$, where $'c'$ is an integer, then $f(0)+f(-1)$ is

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

  1. an even integer

  2. an odd integer always disable by $3$

  3. an odd integer not divisible by $3$

  4. an odd integer may or not be divisible by $3$

Show answer
Correct Option: D
Explanation:

$f(0) = c$

$f(-1) = c-5$

$f(0)+f(-1) = 2c-5 = 2(c-3) + 1$

As the above is of the form $2k+1$, it is always odd.

For $c=3$, the above is not divisible by 3 but for $c=4$, it is. Therefore, it may or may not be divisible by 3.

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

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

  1. $5014$

  2. $4514$

  3. $5414$

  4. $5114$

Show answer
Correct Option: C
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 

Difference of squares of two odd integers is always divisible by ?

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

  1. 3

  2. 5

  3. 16

  4. 8

Show answer
Correct Option: D
Explanation:
Let two consecutive odd integers be $2k +1$ and $2k+3$, where $k$ is any integer
Difference$={\left(2k+3\right)}^{2}-{\left(2k+1\right)}^{2}$
$=4{k}^{2}+9+12k-4{k}^{2}-1-4k=8k$
Hence, the difference is always divisible by $8$.

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

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

  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 $2705$

Show answer
Correct Option: B,C
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.

The smallest odd number formed by using the digits $1,0,3,4$ and $5$ is

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

  1. $10345$

  2. $10453$

  3. $10543$

  4. $10534$

Show answer
Correct Option: A
Explanation:

The smallest odd number using digits $1,0,3,4,5$


$\rightarrow $ We have five digits and we have to make smallest five digit odd numbers.


$\rightarrow$ So, the number cannot start with $0$

$\rightarrow$ For the smallest it should be start with $1$

$\rightarrow$ and second space should be $0$

    $1\\ \overline { 1st } $  $0\\ \overline { 2nd }$  $\;\\ \overline { 3rd } $  $\;\\ \overline { 4rt } $  $\;\\ \overline { 5th } $

$\rightarrow$ Now two space are filled and $3$ are left.

$\rightarrow$ For smallest third place for should be $3$ 

          $\underline { 1 } \underline { 0 } \underline { 3 } \underline {  } \underline {  } $

$\rightarrow $ Now two places are left for and no. should be odd so, last digit should be $5$

So, the number $=10345.$

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

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

  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

Show answer
Correct Option: A

Total number of four digit odd numbers that can be formed using $0,1,2,3,5,7$ are

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

  1. $192$

  2. $375$

  3. $400$

  4. $720$

Show answer
Correct Option: D
Explanation:

we have the number $0,1,2,3,5,7$

Now the digit should be odd and hence last digit should be 
filled with an odd number 
$ \Rightarrow $ Number of way to filled last number $= 4$ $({\text{i}}{\text{.e}}{\text{. }}1,3,5,7)$
$ \Rightarrow $ Number of way to filled third digit  $= 6$ $({\text{i}}{\text{.e}}{\text{. 0,}}1,2,3,5,7)$
$ \Rightarrow $ Number of way to filled second digit $= 6$ $({\text{i}}{\text{.e}}{\text{. 0,}}1,2,3,5,7)$
$ \Rightarrow $ Number of way to filled first digit $= 5$ $({\text{i}}{\text{.e}.}1,2,3,5,7)$
$ \Rightarrow $ Total 4digit number $=4\times6\times6\times5$
$= 720$
hence,
Opton $D$ is correct answer.

The number of even proper divisor of 1008 is

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

  1. 18

  2. 17

  3. 23

  4. 9

Show answer
Correct Option: C
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.

The product of two odd numbers is

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

  1. An even numbers

  2. An odd number

  3. Cannot be determined

  4. None of these

Show answer
Correct Option: B
Explanation:

An odd number is an integer which is not a multiple of two. For example $1,3,5,7,9,11......$


Let us take two odd numbers $a=3$ and $b=5$ and the product of $a$ and $b$ is as follows:

$a\times b=3\times 5=15$

Since $15$ is not a multiple of two, therefore, $15$ is also an odd number.

Hence, the product of two odd numbers is an odd number.

A, Band C are three consecutive even intergers such that three times the first is two more the twice the third one. What is third one?

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

  1. 11

  2. 12

  3. 14

  4. 10

Show answer
Correct Option: C
Explanation:

Let the first even consecutive integer be $x$ then the next two integers would be $x+2$ and $x+4$.


It is given that three times the first number is two more than twice the third number, therefore, we have:

$3x=2(x+4)+2\ \Rightarrow 3x=2x+8+2\ \Rightarrow 3x=2x+10\ \Rightarrow 3x-2x=10\ \Rightarrow x=10$

Thus, the third integer will be $x+4=10+4=14$.

Hence, the third integer is $14$.