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 April 7th, 2013, 08:15 AM #1 Member   Joined: Jan 2013 Posts: 47 Thanks: 0 Product of prime factors Hi I am stuck on the question: What is the smallest square number that is a multiple of 252 You need to use what you got in your answer to the first part of the question Which was prime factor decomposition of 252 in which I got 2 squared times 3 squared times 7 Could someone tell me how would you work this out
 April 7th, 2013, 08:36 AM #2 Member   Joined: Apr 2013 Posts: 65 Thanks: 0 Re: Product of prime factors $252= 2^2 \cdot 3^2 \cdot 7^1$ Square numbers can be written like this: $p_{1}^{2k_{1}} \cdot p_{2}^{2k_{2}} \cdot p_{3}^{2k_{3}} \cdot \cdot \cdot p_{n}^{2k_{n}}$ Because the power of the 7 is odd, we have to multiply it with 7. So the answer is: $\Rightarrow 2^2 \cdot 3^2 \cdot 7^2= 42^2$

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