Highest Common Factor (HCF) and Greatest Common Divisor (GCD)

$348=$$2\times$$2\times$$3\times$$29$

HCF of 144 and 198 is:

$ 198 = 2 \times 3^2 \times 11$
$ 144 = 2^4 \times 3^2$

Highest Common factor is 9.
Option A is correct

Factors of the given numbers are,

Let's look at products of $6$
$6, 12, 18, 24, 30, 36, 42, 48, 54, 60, 66, 72, 78, 84, 90$
The pairs whose sums equal to $90$ are:
$6 + 84=90$
$12 + 78=90$
$18 + 72=90$
$66 + 24=90$
$60 + 30=90$
$54 + 36=90$
$48 + 42=90$
Total number of pairs are $7$.
But there can be  $7$ more pairs when the numbers are reversed.
$\therefore$ total number of ordered pair are $14$

$x= 2^3 \times 3 \times 5^2$
$y = 2^2 \times 3^3$
HCF (x,) $=2^2 \times 3$= 12

Pair in Option C is not possible.
because in option C pair is $(42,80)$
Since, Product of numbers $=42\times 80 =3360$
Option C is correct.

Since, $(a + b -c)^6 = (a + b -c)^4 \times (a + b -c)^2 $
$\therefore$  G.C.D. of $(a + b -c)^6$ and $(a + b -c)^4$ = $(a + b -c)^4$
$\because (a + b -c)^4$ is greatest common in $(a + b -c)^4$ and $(a + b -c)^6$.
Option D is correct.

Let, $p(x) = 20x^2-9x+1$ and $q(x) = 5x^2-6x+1$
$p(x) = 20x^2-9x+1$
         $=20x^2-5x-4x+1$
         $=5x(4x-1)-1(4x-1)$
         $=(5x-1)(4x-1)$
and
$q(x) = 5x^2-6x+1$
         $=5x^2-5x-x+1$
         $=5x(x-1)-1(x-1)$
         $=(5x-1)(x-1)$
$\therefore$ G.C.D of $p(x)$  and  $q(x)=(5x-1)$.
Option B is correct.

Let $p(x) = x^3+x^2-x-1$ and $ q(x) = x^2-1$
$p(x) = x^3+x^2-x-1$
         $ = x^2(x+1)-1(x+1) $
         $ = (x^2-1) (x+1) $
         $ = (x-1)(x+1)(x+1) $
and
$ q(x) = x^2-1$
         $= (x+1)(x-1) $
$\therefore $ G.C.D of $p(x)$ and $q(x)$ =$ (x+1)(x-1) = x^2 - 1 $
Option A is correct.

$p(x) = (x^2-9)(x-3)$
        $= (x^2-3^2)(x-3)$
        $= (x-3)(x+3)(x-3)$
and
$q(x) = x^2+6x+9$
        $ = x^2+3x+3x+9$
        $ = x(x+3)+3(x+3)$
        $= (x+3)(x+3) $
$\therefore $ G.C.D of $p(x)$ and $q(x) = x+3 $
Option C is correct.

$p(x) =8(x^4-16)$
   $= 4\times 2 [(x^2)^2 - 4^2] $
   $ = 4\times 2 (x^2 - 4) (x^2 + 4) $
   $ = 4\times 2  (x+2)(x-2) (x^2 +4) $
and
 $q(x) = 12(x^3-8)$
       $=4\times 3 (x^3 - 2^3) $
         $=4\times 3 (x - 2)(x^2 + 2x + 4) $
$\therefore $ G.C.D of $p(x)$ and $q(x) = 4(x-2) $
Option A is correct.

${ 3 }^{ 5 }$ = 3x3x3x3x3

Let the two numbers be $a$ and $b$.
$a=quotient\times dividend$
$a=21\times 9$
  $=189$
$\therefore b=\frac { GCD\times LCM }{ a }$
                  $=\frac { 27\times 2079 }{ 189 }$
                  $=297$

$37-1=36, 50-2=48$
$123-3=120$
HCF of 36, 48 and $120=12$
$\therefore$ required number $=12$.

$1134 = 2\times3^{4}\times7$

$10=5\times 2$
$100=5\times 2\times 5\times 2=5^2\times 2^2$
$HCF = 5\times 2=10$

The factors of the given numbers are :
$45  = 1,3,5,9,15$ and $45$
$135 = 1,3,5,9,15,45$ and $135$
$270 = 1,3,5,9,15,45,90,135$ and $270$
The common factors in each of the above numbers are $3,3$ and $5$.
Hence, the GCF is $3\times 3\times 5$ = 45
and as $45$ is also a factor of both $135$ and $270$.
The GCF of $45, 135$ and $270$ is $45$.
The correct answer is Option E, the number $45$.

Ans: E

We know, $42=2\times 3\times 7$

$4$ positive integers are factors of $28$ and $42$ both.
For example,
$1,4, 7, 6$ and $28$ itself is a factor of $28$.
$1,4, 7, 6$ and $42$ itself is factor of $42$.

In the given answers $30$ is a factor of $60$, as factors of $60$ are
$2,3,4,5,6,10,12,15,30,20,60$

Given numbers are $1.08$, $0.36$ and $0.90$.

Let the numbers be $3x$, $4x$ and $5x$.
Then, their L.C.M. $= 60x$.
So, $60x = 2400$ or $x = 40$.
$\therefore $ The numbers are $\left( 3\times 40 \right) $, $\left( 4\times 40 \right) $ and $\left( 5\times 40 \right) $.
Hence, required H.C.F. $= 40$.

Required H.C.F. $=\dfrac { H.C.F.\quad of\quad 9,12,18,21 }{ L.C.M.\quad of\quad 10,25,35,40 } = \dfrac { 3 }{ 1400 } $

The given terms are $\dfrac {3}{1}, \dfrac {6}{5}$ and $\dfrac {3}{50}$.
H.C.F. of these terms $= \dfrac {\text{H.C.F. of}\ 3, 6, 3}{\text{L.C.M. of}\ 1, 5, 50}$
$= \dfrac {3}{50} = 0.06$

$x^{4} - 3x + 2 = x^{4} - x^{3} + x^{3} - x^{2} + x^{2} - x - 2x + 2$
$= x^{3} (x - 1) + x^{2} (x - 1) + x(x - 1) - 2(x - 1)$
$= (x - 1) [x^{3} + x^{2} + x - 2]$
$x^{3} - 3x^{2} + 3x - 1 = (x - 1)^{3}$
$x^{4} - 1 = (x - 1) (x + 1) (x^{2} + 1)$
$HCF = x - 1$

$5xy = 15xy$

$28ab = 1722ab$

G.C.D = $1$

We know that

Factorization of the following

Factorization of the following

There is no common factor between $4$ and $19$. Hence, G.C.D. of $4$ and $19$ is $1$.

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

$85 = 5 \times  17$

HCF of $85  and  153 = 17$

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

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

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

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

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

$\Rightarrow$$25=5\times5$

$\Rightarrow$$30=5\times6$

Hence the Highest common factor is 5

Therefore HCF is $5$

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

Given,

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.

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

$ab = 1.ab$
$bc= 1.bc$
$ca = 1.ca$

Since (x + 6) is the HCF, (x + 6) must be a factor of both polynomials. For p(x) = x^2 - a, x = -6 makes p(-6) = 36 - a = 0, so a = 36. Thus p(x) = (x - 6)(x + 6). For q(x) = x^2 - bx + 6, x = -6 makes 36 + 6b + 6 = 0, so 6b = -42, b = -7. Thus q(x) = x^2 + 7x + 6 = (x + 6)(x + 1). The ratio p(x)/q(x) simplifies to (x - 6)/(x + 1).

$(a+b)^2= 1 (a+b)(a+b)$
$(a-b)^2= 1 (a-b)(a-b)$
$HCF= 1$

As the HCF is $ 12 $, the numbers can be written as $ 12x $ and $ 12y $, where x and y are co-prime to each other.
So, $ 12x + 12y = 84 => x + y = 7 $

The pair of numbers that are co-prime to each other and sum up to $7$ are $(2, 5), (1,6), (3,4)$.
Hence, only $ 3 $  pairs of such numbers are possible.
 The numbers are $ (24, 60), (12,72) $ and $ (36,48) $

Prime factorisation of $ (x+2)({x}^{2}+8x+15) = (x+2)  \times [(x+3) \times (x+5)] $
Prime factorisation of $ (x+3)({x}^{2}+9x+20)
= (x+3) \times [(x+4) \times (x+5)] $
So,HCF $  =  (x+3) \times (x+5) = ({x}^{2}+8x+15) $

Given, HCF $ = 4x-6 = 2(2x-3) $

Since HCF needs to be a factor of both the polynomials, clearly only option C with polynomials $ 2({2x-3)}^{2} , 4(2x-3) $  have both factors $ 2 $ and $ (2x-3) $

The HCF of $120,144,216$ is

Required rate = H.C.F. of (68 km, 102 km, 51 km) = 17 km per day

The HCF of 136, 170, and 255 is found by prime factorization: 136 = 8 * 17, 170 = 10 * 17, 255 = 15 * 17. The greatest common divisor is 17.

The required number will be the H.C.F of (120 - 5), (165 - 4) and (210 - 3) i.e. H.C.F. of 115
161 and 207
$\displaystyle \therefore $ Required number = H.C.F. of 115, 161 and 207 = 23

The HCF of $24,36,92$ can be found by factorising all three numbers:

$18 = 2 \times 3 \times 3$

$75=5^2\times 3$

$825=3\times5\times5\times 11$

$12 = 2 \times 2 \times 3 $

Factorization of the following.

A number divides 926 and 2313 leaving remainders 2 and 3 respectively.

This means that the number perfectly divides 926 - 2 = 924 as well as 2313 - 3 = 2310.

Now we simply need to find the HCF of 924 and 2310 

On calculating the HCF , we get it as  462

Hence, the answer is 462

The difference between each set of two numbers is respectively 18, 42 and 24.
Now, as the remainder should be the same in each case, the number is the greatest common divisor of 18, 24 and 42.
$18=2\times 3\times 3$
$24=2\times 2\times 2\times 3$
$42=2\times 3\times 7$
$\therefore HCF=2\times 3=6$
$\therefore$ The required number is 6.

When two numbers have no common factors other than $1$, so they are co-prime numbers.

To find the HCF of $36$ and $ 144 $, first factorise them: