HCF and LCM

Using Euclidean algorithm: 420 = 130 * 3 + 30; 130 = 30 * 4 + 10; 30 = 10 * 3 + 0. The HCF is 10.

The given number $100$ and $101$ are consecutive numbers and respectively even and odd numbers.

  HCF of 210 and 55 is 5

HCF of $468$ and $222$
$468 = \left(222 \times 2\right) + 24$
$222 = \left(24\times\ 9\right) + 6$
$24 = \left(6\times\ 4\right) + 0$
$\therefore HCF = 6$

Given the H.C.F. of $A$ and $B$ is $24$ and that of $C$ and $D$ is $56.$
Then the H.C.F. of $A, B, C$ and $D$ is the HCF of $24$ and $56$ which is $8.$

136)170(1

$480=2^5\times 3\times 5$
$720=2^4\times 3^2\times 5$
$108=2^2\times 3^2$
$HCF=2^2\times 3=12$

$248=2^3 \times 31$

$26=13\times 2$

$1650 = 2\times 3\times 5^{2}\times 11$
$90 = 2\times 3^{2} \times 5^{1}$
$1500 = 2^{2} \times 3^{1} \times 5^{3}$
H.C.F. of $1650, 90$ and $1500$ is $2\times 3\times 5 = 30$

$HCF$ of two consecutive even numbers is always $2$. For example:

Factors of 6 are $1,2,3,6$
Factors of 72 are $1,2,3,4,6,8,9,12,18,24,36,72$
Factors of 120 are $1,2,3,4,5,6,8,10,12,15,20,24,30,40,60,120$
The highest factor in all the three is 6 .