# Express 540 as a product of its prime factor?

Unfortunately, I can’t reply to you with a picture, but if you know what a prime factor tree is, that’s a helpful way to work through each step visually. This is how you’d do one for the number 16: https://socratic.org/questions/what-is-a-prime-factor-tree-of-16 The fact that 2 goes into 540 would be a good place to start! Hope that helps!

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

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