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### 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
```

```
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

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/
```
Associated Topics:
Middle School Factoring Numbers
Middle School Prime Numbers

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