SOLUTION: what is the smallest number that is divisible by 585 nd 624 ?

Algebra ->  Divisibility and Prime Numbers -> SOLUTION: what is the smallest number that is divisible by 585 nd 624 ?       Log On


   

Question 972502: what is the smallest number that is divisible by 585 nd 624 ?
Found 2 solutions by MathLover1, Edwin McCravy:
Answer by MathLover1(20855) About Me  (Show Source):
You can put this solution on YOUR website!
recall:
Least Common Multiple - The smallest number that is divisible by two or more given numbers.
so, find LCM
585|5+
117|3
39|3
13|13
1
=>585=3%5E2%2A5%2A13
and
624|2
312|2
156|2
78|2
39|3
13|13
1
=>624=2%5E4%2A3%2A13
LCM=2%5E4%2A3%5E2%2A5%2A13=16%2A9%2A5%2A13=9360
so, the smallest number that is divisible by 585 and 624 is 9360




Answer by Edwin McCravy(20083) About Me  (Show Source):
You can put this solution on YOUR website!
That's called "the least common multiple"or lcm(585,624).
We write the prime factorization of each:

Look at prime factor 2:

585 = 3*3*5*13
624 = 2*2*2*2*3*13

2 is a factor of 585 0 times.
2 is a factor of 624 4 times.

So 2 is a factor of lcm(585,624) 4 times because 4 ≥ 0.
----------------------

Look at prime factor 3:

585 = 3*3*5*13
624 = 2*2*2*2*3*13

3 is a factor of 585 2 times.
3 is a factor of 585 1 time.

So 3 is a factor of lcm(585,624) 2 times because 2 ≥ 1.
----------------------

Look at prime factor 5:

585 = 3*3*5*13
624 = 2*2*2*2*3*13

5 is a factor of 585 1 time.
5 is a factor of 585 0 times.

So 5 is a factor of lcm(585,624) 1 time because 1 ≥ 0.
----------------------

Look at prime factor 13:

585 = 3*3*5*13
624 = 2*2*2*2*3*13

13 is a factor of 585 1 time.
13 is a factor of 585 1 time.

So 13 is a factor of lcm(585,624) 1 time because 1 ≥ 1.
----------------------

Put it all together:

 2 is a factor of lcm(585,624) 4 times. 
 3 is a factor of lcm(585,624) 2 times.
 5 is a factor of lcm(585,624) 1 time.
13 is a factor of lcm(585,624) 1 time.

So lcm(585,624) = 2*2*2*2*3*3*5*13 = 9360. 

Edwin