Use the euclidean algorithm to find the gcd of 2232 and 3828 and find the lcm of 2232 and 3828.
I'm not expecting an answer maybe just a quick and easy example of how to solve the sum quickly..thanks
Use the following theorem,Originally Posted by kodirl
$\displaystyle \gcd(a,b)\mbox{lcm} (a,b)=ab$
Procede with Euclidean Algorithm,
$\displaystyle 3828=(1)2232+1596$
$\displaystyle 2232=(1)1596+636$
$\displaystyle 1596=(2)636+324$
$\displaystyle 636=(1)324+312$
$\displaystyle 324=(1)312+12$
$\displaystyle 312=(26)12+0$
Therefore,
$\displaystyle \gcd(3828,2232)=12$
Therefore,
$\displaystyle \mbox{lcm}(3828,2232)=\frac{3828\cdot 2232}{12}=712008$
Hello, Kerry!
Do you know the Euclidean Algorithm?
If you do, exactly where is your difficulty?
Use the Euclidean Algorithm to find the GCD and LCM of 2232 and 3828.
Example: Find the GCD of 72 and 120.
Step 1: Divide the larger by the smaller.
. . . . . . $\displaystyle 120 \div 72 \:=\:1,\;rem.\,48$
Step 2: Divide the divisor by the remainder.
. . . . . . $\displaystyle 72 \div 48\:=\:1,\;rem.\,24$
Repeat Step 2 until a zero remainder is achieved.
. . . . . . $\displaystyle 48 \div 24\:=\:2,\;rem.\, 0$
. . . . . . . . . .$\displaystyle \uparrow$
. . The last divisor is the GCD.
Therefore: .$\displaystyle GCD(72,\,120)\:=$ 24
There is a formula for the LCM: .$\displaystyle \boxed{\boxed{LCM(a,b)\;=\;\frac{a\cdot b}{GCD(a,b)}}} $
Hence: .$\displaystyle LCM(72,120)\;=\;\frac{72\cdot120}{24}\;=$ 360