What is the smallest three digit number divisible by the first three prime numbers and the first three composite numbers?

Question

anonymous · Answer

$$2\times3\times5\times2\times2= 120$$

anonymous · Answer

How did you get it ?

anonymous · Answer

2, 3 and 5 are first three prime numbers. 2, 3 and 5 are its factors. and now it is also divisible by first three composite number. first three composite numbers are 4, 6 and 8. 

for divisible by 4 it should have 2 X 2 as factor. since one 2 is already there i multiplied one more 2. 

for divisible by 6 it should have 2 X 3 as factor and its already there. I dont need to multiply anything more. 

for divisible by 8 it should have 2 X 2 X 2 since two 2's are already there I multiplied with one more 2. 

and thats how I got the answer.

mathstudent55 · Answer

Here's another way of doing it.

The first three prime numbers are 2, 3, 5.
The first three composite numbers are 4, 6, 8.

This problem is actually asking for the least common multiple (LCM) of 2, 3, 4, 5, 6, 8.

Let's do the prime factorization of all 6 numbers:

$2 = 2$
$3 = 3$
$4 = 2^2$
$5 = 5$
$6 = 2 \times 3$
$8 = 2^3$

Rule:
To find the LCM of several numbers, first list the numbers' prime factors. Then pick common and not common factors with the highest exponent. The LCM is the product of these picked factors.

We see the common factor of 2 with exponents 1, 2, and 3. Pick $2^3$, the highest exponent on a 2.
Then we see not-common factor of 3. Pick it, too.
The we see not-common factor of 5. Pick it, too.

In the end, we have picked $2^3$, 3, and 5.

LCM = $2^3 \times 3 \times 5 = 120$

anonymous · Answer

Thanks!

mathstudent55 · Answer

You're welcome.