Tag: common factors and hcf

Questions Related to common factors and hcf

The HCF of $2{x^2}$ and $12{x^2}$ is

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

  1. $2{x^2}$

  2. $12{x^2}$

  3. $2x$

  4. $12x$

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Correct Option: A
Explanation:

$2x^2=2\times x\times x$


$12x^2=6\times 2\times x\times x$

          $=3\times 2\times 2 \times x\times x$

Common factor between $2x^2$ and $12x^2=2\times x\times x=2x^2$

$\therefore$  H.C.F of $2x^2$ and $12x^2$ is $2x^2$.

Find the HCF of $25$ and $30$.

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

  1. $25$

  2. $6$

  3. $1$

  4. $5$

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Correct Option: D
Explanation:

$\Rightarrow$$25=5\times5$

$\Rightarrow$$30=5\times6$

Hence the Highest common factor is 5

Therefore HCF is $5$

The solution of: $8\mod x\equiv 6\mod 14$ is,

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

  1. ${8, 6}$

  2. ${6, 14}$

  3. ${6, 13}$

  4. ${8, 14}$

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Correct Option: C
Explanation:
Solution:-
$8x \equiv 6 \left( mod \ 14 \right)$

$\because \; gcd \left( 8, 14 \right) = 2 \text{ divides } 6$

To find solutions, we first solve

$8x − 14y = 6$

By trial and error method, we find a solution

$\left( x, y \right) = \left( 6, 3 \right)$

This means that $x \equiv 6 \left( mod \ 14 \right)$ is a solution

To the congruence $8x \equiv 6 \left( mod \ 14  \right)$

$\therefore$ Incongruent solutions are,

$x = 6 +\left ( k \times \dfrac{14}{2} \right );  k = 0, 1$

$\therefore \; x = 6, 13$

Hence option $C$ is the answer.

If $G.C.D\ (a , b) = 1$ then $G.C.D\ ( a+b , a-b )$=?

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

  1. $1$ or $2$

  2. $a$ or $b$

  3. $a+b$ or $a-b$

  4. $4$

Show answer
Correct Option: A
Explanation:
It is given that GCD$\left(a,b\right)=1$

Let GCD$\left(a-b,a+b\right)=d$

$\Rightarrow\,d$ divides $a-b$ and $a+b$

there exists integers $m$ and $n$ such that 

$a+b=m\times d$        ..........$(1)$

and $a-b=n\times d$        ..........$(2)$

Upon adding and subtracting equation $(1)$ and $(2)$ we get

$2a=\left(m+n\right)\times d$         ..........$(3)$

and $2b=\left(m-n\right)\times d$         ..........$(4)$

Since, GCD$\left(a,b\right)=1$(given)

$\therefore\,2\times GCD\left(a,b\right)=2$

$\therefore\,GCD\left(2a,2b\right)=2$ since $GCD\left(ka,kb\right)=kGCD\left(a,b\right)$

Upon substituting  value of $2a$ and $2b$ from equations $(3)$ and $(4)$ we get

$\therefore\,gcd\left(\left(m+n\right)\times d,\left(m-n\right)\times d\right)=2$

$\therefore\,d\times gcd\left(\left(m+n\right),\left(m-n\right)\right)=2$

$\therefore\,d\times$ some integer$=2$

$\therefore\,d$ divides $2$

$\therefore\,d\le 2$ if $x$ divides $y,$ then $\left|x\right|\le \left|y\right|$

$\therefore\,d=1$ or $2$ since, gcd is always a positive integer.

ILLUSTRATION  2 The total number of factors (exculding 1) of 2160 is 

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

  1. 40

  2. 39

  3. 41

  4. 38

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Correct Option: A

The GCD of two numbers is $17$ and their LCM is $765$. How many pairs of values can the numbers assume?

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

  1. $1$

  2. $2$

  3. $3$

  4. $4$

Show answer
Correct Option: B
Explanation:

Since the GOD of numbers is $17$. So, the numbers are $17a$ and $17b$, where a and b are relatively prime.
LCM$=765$

$\Rightarrow 17a\times 17b=765$
$\Rightarrow ab=45$
$\Rightarrow a=15, b=9$ or $a=9$, $b=5$.
So, the numbers are $17\times 5=85$ and $17\times 9=153$.
The numbers can be $17\times 1=17$ and $765$. So, two pairs are possible.

If two positive integers $a$ and $b$ are written as $a=x^3y^2$ and $b=xy^3$; $x, y$ are prime numbers, then HCF of $a$ and $b$ is

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

  1. $xy$

  2. $xy^2$

  3. $x^3y^3$

  4. $x^2y^2$

Show answer
Correct Option: B
Explanation:

Given,

$a={  x}^{3  }{ y }^{2  } = x\times x\times x\times y\times y$

$b={  x}{ y }^{3  }         =x\times y\times y\times y$

H.C.F of $a,b$  = ${  x}{ y }^{2  } $

When teams of same size are formed from three groups of $512, 430$ and $489$ students separately $8, 10$ and $9$ students respectively are left out What could be the largest size of the team?

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

  1. $6$

  2. $12$

  3. $18$

  4. $20$

Show answer
Correct Option: B
Explanation:

It is given that $8,10$ and $9$ students are respectively left out from the three separate groups $512,430$ and $489$ when the teams of same size are formed.


Number of students taken from first group are $512-8=504$
Number of students taken from second group are $430-10=420$
Number of students taken from third group are $489-9=480$

Now, we factorize $504,420$ and $480$ as follows:

$504=2\times 2\times 2\times 3\times 3\times 7\ 430=2\times 2\times 3\times 5\times 7\ 480=2\times 2\times 2\times 2\times 2\times 3\times 5$

Therefore, the HCF of $504,420$ and $480$ is:

HCF$\left( 504,430,480 \right) =2\times 2\times 3=12$

Hence, the largest size of the team is of $12$ students.

HCF of the numbers $24,36$ and $92$ is 

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

  1. $24$

  2. $36$

  3. $12$

  4. $4$

Show answer
Correct Option: D
Explanation:

Factors of $ 24 =1,2,3,4,6,8,12,24 $
Factors of $36= 1,2,3,4,6,9,12,18,36$

Factors of $ 92=1,2,4,23,46,92 $

Common factors are $ =1,2,4 $
$\therefore $ HCF $=4.$

If HCF of $m$ and $n$ is $1,$ then what are the HCF of $m + n, m$ and HCF of $m - n, n$ respectively? 

$\displaystyle \left ( m> n \right )$

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

  1. $1$ and $2$

  2. $2$ and $1$

  3. $1$ and $1$

  4. cannot be determined

Show answer
Correct Option: C
Explanation:
Let us consider an example.
Let $m =16$ and $n =9$ be relatively prime numbers.
So, $m+n=25$. The HCF of $25$ and $16$ is $1$. 

$m-n=7$. The HCF of $7$ and $9$ is $1$.
Similarly, if we take other values for $m$ and $n,$ we get the same answer. 
Therefore, option $C$ is correct.