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Asked by Lacey

Express 540 as a product of its prime factor?

A prime factor tree would be a good method to use here. First think of 2 numbers that multiply to get 540, the most obvious ones are 54 and 10. Now both of these are not prime numbers. So we'll look at 10 first, which is a product of 5 and 2 and both of these are primes so we are finished with 10. Now looking at 54, we can see that 9 and 6 are factors (as well as 2 and 27, 3 and 18). Both 9 and 6 are not primes so we look at 6 which has factors 3 and 2. Looking at 9 its factors are 3 and 3. Now we collect all the primes and we have: 5,2,3,2,3,3 so 54= 5 * 2^2 *3^3

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Zahra Bashir

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A prime number is one which does not divide into multiples of smaller numbers and are all odd numbers so not divisible by 2. For 540 you start by divide it by the smallest prime number as many times until no more 2 is available ie (540/2)=270 then (270/2) = 135 then this odd so divide by 3 (135/3) = 45. again odd so divide by 3 (45/3)= 15 repeat this (15/3)=5 so now you stop no more division So 540= 2x2x3x3x3x5 These are the prime cofactors or rewritten 2^2x3^3x5^1 Or 2^2x3^3x5

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A prime number is one which does not divide into multiples of smaller numbers and are all odd numbers so not divisible by 2. For 540 you start by divide it by the smallest prime number as many times until no more 2 is available ie (540/2)=270 then (270/2) = 135 then this odd so divide by 3 (135/3) = 45. again odd so divide by 3 (45/3)= 15 repeat this (15/3)=5 so now you stop no more division So 540= 2x2x3x3x3x5 These are the prime cofactors or rewritten 2^2x3^3x5^1 Or 2^2x3^3x5

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