Least Common Multiple, 2-10

Date: 10/28/2002 at 21:30:55
From: Beatrice Ambler
Subject: Test for divisibility

I'm taking the course "Mathematics for Elementary Teachers." One 
question in our homework assignment is to find the smallest counting 
number that is divisible by the numbers 2 through 10.

I began by trying to test for divisibility for the units digit. By 
saying that a number is divisible by 2, 5, and 10 if its units digit 
is 0. I moved next to the tens digit and tried testing for a number 
divisible by 4 and 8. Then I moved to the hundreds place and looked 
for a number divisible by 3 and 9. I reached 720, which is divisible 
by all the numbers 2 through 10 except 7. The answer key shows the 
answer to be 2520. What would be the logical steps to follow in 
reaching this answer?

Date: 10/28/2002 at 22:01:08
From: Doctor Paul
Subject: Re: Test for divisibility

I think the best way to do this problem is just to construct a number 
that meets the described property:

In order to be divisible by two, we need a two in its prime 
factorization.  So our number starts out as:

   2

next we need the number to be divisible by three, so a three must 
appear in its prime factorization. Thus our number has become:

   2*3

Now we want our number to be divisible by four. So we need two twos 
in its prime factorization. We already have one two. So we need to 
add one more:

   2*3*2

Continuing in a similar pattern will establish the fact that
2*3*2*5*7*2*3 = 2520 is the smallest natural number evenly divisible 
by each of the integers two through ten.

I hope this helps.  Please write back if you'd like to talk about 
this some more.

- Doctor Paul, The Math Forum
  http://mathforum.org/dr.math/