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Find the Smallest Number...


Date: 10/21/97 at 21:42:26
From: Cheryl Starzyk
Subject: Help

Can you show me a example of how to do this math problem? 

I am the smallest number that has factors of 1, 2, 3, 4, 5, 6, 7, 
and 8. What number am I?
 
Thank you,
Chad


Date: 10/22/97 at 17:21:31
From: Doctor Chita
Subject: Re: Help

Hi Chad:

A problem in number theory! How interesting! Did you know that this 
problem is similar to what cryptologists use to create secret codes? 
That is, they take prime numbers and multiply them together to create 
a very large number that is the secret to a cipher, or code. Since 
it's nearly impossible to factor a very large number into primes, the 
code is impossible to decipher.

Anyway, about your problem - think about prime and composite numbers 
for a second. A counting number is either a composite number or prime. 
Composite numbers are made up of prime factors (also called divisors).  
A prime number has only two factors (divisors): itself and 1. 
Therefore, 3 is prime, and 4 is not.

The numbers in your sample include prime and composite numbers. What I 
would suggest is that you rewrite the composite numbers as products of 
their prime numbers. Here's an example.

Suppose my numbers are 2, 5, 6, 12, 21, 24, and 27. I want the 
smallest number having these numbers as divisors.

Here are all the factors: I've put parentheses around the prime 
factors of the composite numbers.

  2 * 5 * (2 * 3) * (2^2 * 3) * (3 * 7) * (2^3 * 3) * (3^3)

Regroup the factors in order:

 [2 * 2 * 2^2 * 2^3] * [3 * 3 * 3 * 3 * 3^3] * 5 * 7

The smallest number I want must contain enough 2s to take care of 
the largest power of 2 in the expression. This is 2^3 = 8. But once 
I have the three 2s, I can take care of the other 2s, since they all 
divide 8. 

I also need to have 3^3 to take care of 27. The remaining 3s also 
divide 27, so I don't need any more of them in my "least" number. 

Since there is only one 7 and one 5, I need one of each in the final 
set.

So my least number would be 

  2^3 * 3^3 * 5 * 7 or 8 * 27  * 5 * 7 = 7560.

Check to be sure that this number can be divided by each of the 
numbers in the original set: 2, 5, 6, 12, 21, 24, and 27.

Now you try it. Rewrite the numbers using exponents and then find the 
largest power of each prime number in your set. The product of primes 
raised to the largest power for each prime will make up the number you 
are looking for.

Isn't math fun!

-Doctor Chita,  The Math Forum
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Associated Topics:
High School Number Theory
Middle School Factoring Numbers

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