Prime Factors and Square Products
Date: 10/05/2003 at 15:59:25 From: Alexander Subject: Creating a square number What is the smallest number that you can multiply by 540 to make a square number? I know the answer is 15, which I found by trial and error. That is, I worked out the prime factors and they are 2^2, 3^2, and 5. Juggling these brings me to 90 x 90 = 8100, which is 540 x 15. What I cannot work out for numbers other than 540 is the correct method of juggling. My working was random and I'd love to be able to apply this to other numbers. Thanks.
Date: 10/05/2003 at 19:20:29 From: Doctor Luis Subject: Re: Creating a square number Hi Alexander, You were on the right track. The prime factorization of 540 is 540 = (2*2)*(3*3*3)*(5) = 2^2 * 3^3 * 5^1 Now, when you square a number, the exponents get multiplied by two. Therefore, the prime factors of a square number appear an even number of times. So, to make the product of 540 and something else a square number, you have to make the prime factors of the product appear an even number of times. You can see that there are already two 2's, so you don't worry about the 2. But there are three 3's and only one 5. That's why you need an extra 3 (so there are four instead of three) and an extra 5 (so you have two instead of just one). Of course, 3*5 = 15, which was your guess. 540 * (3*5) = 2^2 * 3^4 * 5^2 Since the exponents of the prime factors are all even, we can also find what number it squares 540 * 15 = 2^2 * 3^4 * 5^2 = (2 * 3^2 * 5)^2 = (10 * 9)^2 = 90^2 A square number, just like you wanted! Hopefully, you can now apply this method to numbers other than 540. Let us know if you have any more questions. - Doctor Luis, The Math Forum http://mathforum.org/dr.math/
Date: 10/07/2003 at 16:34:42 From: Alexander Subject: Thank you (Creating a square number) Thank you, Doctor Luis, for your help. I had fun working through other numbers and trying out cubes. Alexander
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