<PRE><B>Question 833791</B>
If an integer ends on 0 or 5, then it is divisible by 5. If it ends on 0, 2, 4, 6 or 8, then it is divisible by 2. It is divisible by 3 if the sum of its digits is divisible by three.
     Using these simple rules, determine the prime-factorings of 2295 and 2040. Then use these results to compute the prime-factorings of the following three numbers: 2295 * 2040, gcd(2295,2040) and lcm(2295,2040). Finally, demonstrate the relationship between these three numbers.
&lt;pre&gt;
Prime factorization, &quot;PF&quot;, of 2295 = 3×3×3×5×17 = 3&lt;sup&gt;3&lt;/sup&gt;×5×17 
Prime factorization, &quot;PF&quot;, of 2040 = 2×2×2×3×17 = 2&lt;sup&gt;3&lt;/sup&gt;×3×17

Line them up like this, putting in all factors of both numbers,
When there is a factor of only one of them, include that factor in
the other PF to the 0 power.  Also include any 1 exponents:


     PF(2295) = 2&lt;sup&gt;0&lt;/sup&gt;×3&lt;sup&gt;3&lt;/sup&gt;×5&lt;sup&gt;1&lt;/sup&gt;×17&lt;sup&gt;1&lt;/sup&gt; 
     PF(2040) = 2&lt;sup&gt;3&lt;/sup&gt;×3&lt;sup&gt;1&lt;/sup&gt;×5&lt;sup&gt;1&lt;/sup&gt;×17&lt;sup&gt;1&lt;/sup&gt;
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To get the prime factorization of their product, ADD the two exponents of
each factor and write the result below the line. 

     PF(2295) = 2&lt;sup&gt;0&lt;/sup&gt;×3&lt;sup&gt;3&lt;/sup&gt;×5&lt;sup&gt;1&lt;/sup&gt;×17&lt;sup&gt;1&lt;/sup&gt; 
     PF(2040) = 2&lt;sup&gt;3&lt;/sup&gt;×3&lt;sup&gt;1&lt;/sup&gt;×5&lt;sup&gt;1&lt;/sup&gt;×17&lt;sup&gt;1&lt;/sup&gt;
----------------------------
PF(2295×2040) = 2&lt;sup&gt;3&lt;/sup&gt;×3&lt;sup&gt;4&lt;/sup&gt;×5&lt;sup&gt;2&lt;/sup&gt;×17&lt;sup&gt;2&lt;/sup&gt; = 4681800  

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To get their greatest common divisor, below the line, bring down each 
factor with the MINIMUM exponent: 

      PF(2295) = 2&lt;sup&gt;0&lt;/sup&gt;×3&lt;sup&gt;3&lt;/sup&gt;×5&lt;sup&gt;1&lt;/sup&gt;×17&lt;sup&gt;1&lt;/sup&gt; 
      PF(2040) = 2&lt;sup&gt;3&lt;/sup&gt;×3&lt;sup&gt;1&lt;/sup&gt;×5&lt;sup&gt;1&lt;/sup&gt;×17&lt;sup&gt;1&lt;/sup&gt;
----------------------------
gcd(2295×2040) = 2&lt;sup&gt;0&lt;/sup&gt;×3&lt;sup&gt;1&lt;/sup&gt;×5&lt;sup&gt;1&lt;/sup&gt;×17&lt;sup&gt;1&lt;/sup&gt; = 255

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To get their least common multiple, below the line bring down each 
factor with the MAXIMUM exponent: 


      PF(2295) = 2&lt;sup&gt;0&lt;/sup&gt;×3&lt;sup&gt;3&lt;/sup&gt;×5&lt;sup&gt;1&lt;/sup&gt;×17&lt;sup&gt;1&lt;/sup&gt; 
      PF(2040) = 2&lt;sup&gt;3&lt;/sup&gt;×3&lt;sup&gt;1&lt;/sup&gt;×5&lt;sup&gt;1&lt;/sup&gt;×17&lt;sup&gt;1&lt;/sup&gt;
----------------------------
lcm(2295,2040) = 2&lt;sup&gt;3&lt;/sup&gt;×3&lt;sup&gt;3&lt;/sup&gt;×5&lt;sup&gt;1&lt;/sup&gt;×17&lt;sup&gt;1&lt;/sup&gt; = 18360

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The relationship demonstrated here is that

A×B = gcd(A,B)×lcm(A,B) 

2295×2040 = 4681800 = gcd(2295,2040)×lcm(2295,2040) = 255×18360 = 4681800.

Edwin&lt;/pre&gt;</PRE>