Tag: common factors and hcf

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 $L.C.M.$.