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/
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