Chat with us, powered by LiveChat The factors are 1 2 3 4 6 8 9 12 18 24 36 72. I find factors | Writedemy

The factors are 1 2 3 4 6 8 9 12 18 24 36 72. I find factors

The factors are 1 2 3 4 6 8 9 12 18 24 36 72. I find factors

The factors are 1 2 3 4 6 8 9 12 18 24 36 72. I find factors in pairs It will look like more work than it is because I will explain how I am doing these steps. I do most of the work without writing it down. I’ll put the explanation in black in [brackets] and the answer in ##color(blue)blue##. I’ll proceed by starting with ##1## on the left and checking each number in order until either I get to a number already on the right or I get to a number greater than the of 72. ##color(blue)(1 xx 72)## [I see that 72 is divisible by 2 and do the division to get the next pair] ##color(blue)(2 xx 36)## [Now we check 3 and we get the next pair.] [I use a little trick for this. I know that 36 is divisible by 3 and ##36 = 3xx12##. This tells me that ##72 = 2xx3xx12## so I know that ##72 = 3xx2xx12 = 3xx24##] ##color(blue)(3 xx 24)## [Now we need to check 4. Up above we got ##72 = 2xx36## since ##36 = 2xx18## we see that ##72 = 2xx2xx18 = 4xx18##] ##color(blue)(4 xx 18)## [The next number to check is 5. But 72 is not divisible by 5. I usually write a number before I check so if a number is not a factor I cross it out.] ##color(blue)cancel(5)## {Move on to 6. Looking above I want to ‘build’ a 6 by multiplying a number on the left times a factor of the number to its right. I see two ways to do that: ##2xx36 = 2xx3xx12 = 6xx12## and ##3xx24 = 3xx2xx12=6xx12##. (Or maybe you just know that ##6xx12=72##.)] ##color(blue)(6 xx 12)## [72 is not divisible by 7.] ##color(blue)cancel(7)## {##4xx18 = 4xx2xx9=8xx9##] ##color(blue)(8 xx 9)## [And that’s all. 9 and the factors that are greater than 9 are already written on the right in the list of pairs above.] [Is that clear? Any factor of 72 greater than 9 must be multiplied by something less than 8 to get 72. But we’ve checked all the numbers up to and including 8. So we’re finished.] [If we were doing this for ##39## we would get ##1xx39## and ##3xx13## then we cross off every number until we notice that ##7xx7 = 49##. If 39 had a factor greater than 7 it would have to be multiplied by something less that 7 (otherwise we get 49 or more). So we would be finished.]

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