text: # Finding the smallest positive integer with prime factors.
  
  • Oct 11th 2010, 08:34 AM
  weldon29
  Finding the smallest positive integer with prime factors.
  First I would like to make it clear that I'm not asking this to help with my school work, as I don't go to school and I'm self-studying. So I would appreciate it if people would help me.
  
  So I just started doing some past examination questions and there are a few questions that I'm not sure if I'm using the right method as it took me a long time to finish them.
  Example:
  The numbers 168 and 324, written as the products of their prime factors, are 168 = 2^3 x 3 x 7, 324 = 2^2 x 3^4.
  Find
  the smallest positive integer value of n for which 168n is a multiple of 324.
  ------------------
  Expressed as the product of prime factors, 198 = 2 x 3^2 x 11 and 90 = 2 x 3^2 x 5. Use these results to find
  the smallest integer, k, such that 198k is a perfect square.
  Question ens.
  For the second question, I don't even know how to solve it, as my book didn't really explain it fully.
  • Oct 11th 2010, 08:49 AM
  Pim
  So, for the first question, what you want is the smallest number, which prime factorisation contains both the factors of 168 and the factors of 324. (So, for every factor, you take whichever of the two numbers has a higher power for it and merge them)
  this gives 2^3 (from 168) * 3^4 (from 324) * 7 (from 168) gives 4536. Do you understand my reasoning and why this always works?
  
  For the second question, try and take the square root of a prime factorisation. What condition is true if and only if the number factorised is a perfect square?
  • Oct 11th 2010, 06:50 PM
  weldon29
  Why do I have to multiply with 7 if it is not the highest power? Or am I suppose to take the two highest power if there are more than two numbers to choose from?
  And I still don't understand how to solve the second question. When you say take the square root of a prime factorization, do you mean take the square root of 3^2? And I don't really understand your last sentence either.
  • Oct 11th 2010, 11:26 PM
  Pim
  You can view the factorisation as following:
  168 = 2^3 x 3^1 x 7^1, 324 = 2^2 x 3^4. What I meant by taking the highest power is that for every number in one or both of the factorisations (in this case, 2, 3, 7) you raise this number to the highest power in either of the two (168, 324) numbers.
  I'll demonstrate why 2^3 * 3^4*7 is the correct answer. The best way to do so is divide by the prime factorisations of the two numbers.
  $\frac{2^3*3^4*7}{2^3*3^1*7^1} = 3^3$
  $\frac{2^3*3^4*7}{2^2*3^4} = 2^1*7$
  
  Do you see the pattern and do you understand why the algebra is valid?
  
  As for the second problem
  
  What I meant by taking the square root of a factorisation is that you take the root of the whole factorisation. e.g. \sqrt{198} = \sqrt{2 * 3^2 * 11}
  I'll give you the basis from which you can deduce "what condition must be true" for a number to be a perfect square.
  - Taking the square root of a number is the same as raising it to the power 1/2
  - $(a^p*b^q)^c = a^{pc}*b^{qc}$
  - A number is integer if and only if it is represented by a prime factorisation with integer powers.
  
  If you find what condition must be true, it should be possible to find the answer to your second question. Don't hesitate to ask if you really don't get it.
  • Oct 15th 2010, 09:19 PM
  weldon29
  Sorry for not replying, but I seriously still don't understand so I was kinda avoiding looking at this page.(the only thing I really understood was taking the highest power thing.)
  For one what is the pattern you mentioned? And what is the connection between 3^3 and 2^1 * 7 to 4536?
  And I still don't understand the second question. :(
  • Oct 16th 2010, 12:08 AM
  Pim
  The pattern is that 3^3 and 2*7 do not have any common divisors. Perhaps you should try and read this, Fundamental theorem of arithmetic - Wikipedia, the free encyclopedia
  
  For the second question, lets try an example. 6084 is a perfect square. It's prime factorisation is $2^2*3^2*13^2$
  Let's try taking the square root of that.
  $\sqrt{2^2*3^2*13^2}$
  $(2^2*3^2*13^2)^{\frac{1}{2}}$
  
  As per the formula in my post above:
  $2^1*3^1*13^1$
  
  Do you see why for a number to be a perfect square, the powers of the primes in the prime factorisation have to be even?
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