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

If the HCF of 85 and 153 is expressible in the form 85n $-$ 153, then value of n is :

  1. 3

  2. 2

  3. 4

  4. 1
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

HCF of $85\  and\  153 = 17$

Now given HCf can be expressed in the gorm of $85n-153$
So $17=85n-173$
On solving the above equation we get $n=2$
So correct answer will be option B

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

Choose the correct answer form the alternatives given.
What is the HCF of $(x^4 \, - \, x^2 \, - \, 6) \, and \, (x^4 \, - \, 4x^2 \, + \, 3)$? 

  1. $x^2$ - $3$
  2. $x + 2$
  3. $x + 3$
  4. $x^2$ + $3$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\displaystyle (x^4 \, - \, x^2 \, - \, 6) \, = \, (x^2 \, - \, 3) (x^2 \, + \, 2)$
$\displaystyle (x^4 \, - \, 4x^2 \, + \, 3) \, = \, (x^2 \, - \, 3) (x^2 \, - \, 1)$
HCF is = $x^2$ - $3$

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

Three bells, toll at intervals of $36$ sec, $40$ sec and $48$ sec respectively. They start ringing together at particular time. They next toll together after

  1. $6$ minutes
  2. $12$ minutes
  3. $18$ minutes
  4. $24$ minutes
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

G.C.D of $36,40,48=720\Rightarrow 720sec=12min$
$\therefore$ Next time when three balls toll together is after $12$ mins

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

The G.C.D. of two whole numbers is $5$ and their L.C.M. is $60$. If one of the numbers is $20$, then the other number would be

  1. $23$
  2. $13$
  3. $16$
  4. $15$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

If we are given two numbers $N _1$ and $N _2$ and their $G.C.D$ and $L.C.M.$.

then by property of numbers $N _1$$\times$$N _2=G.C.D$ $\times$ $L.C.M.$

Here Given:
$N _1=20$
$G.C.D.=5$
and $L.C.M=60$
Let, $N _2=x$

then from  above relation
$20$$\times$$x=5$$\times$$60$

$=>x=\dfrac{300}{20}$

$=>x=15$

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

The HCF of $2{x^2}$ and $12{x^2}$ is

  1. $2{x^2}$
  2. $12{x^2}$
  3. $2x$
  4. $12x$
Reveal answer Fill a bubble to check yourself
A Correct answer
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 $2x^2$.

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

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

  1. ${8, 6}$
  2. ${6, 14}$
  3. ${6, 13}$
  4. ${8, 14}$
Reveal answer Fill a bubble to check yourself
C Correct answer
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.
Multiple choice maths hcf-lcm common factors and hcf hcf highest common factor (h.c.f.)

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

  1. $1$ or $2$
  2. $a$ or $b$
  3. $a+b$ or $a-b$
  4. $4$
Reveal answer Fill a bubble to check yourself
A Correct answer
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.