Tag: common factors and hcf

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. $= 40$.