The prime factorization of 540 is 2 x 2 x 3 x 3 x 3 x 5. How do you find the number of divisors of 540 other than 1?

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Answer 1 · 2017-12-06T19:39:36

#540# has #23# positive integer factors apart from #1#

Any positive divisor of #540# is expressible in the form:

#2^a * 3^b * 5^c#

where:

#a in { 0, 1, 2 }#

#b in { 0, 1, 2, 3 }#

#c in { 0, 1}#

So there are #3# possible values of #a#, #4# of #b# and #2# of #c#, giving a total of:

#3 * 4 * 2 = 24#

possible factors. One of those #24# factors is #2^0 * 3^0 * 5^ 0 = 1# which we do not wish to count, leving #23# factors apart from #1#.