Question

find the number of divisors of 720 (including 1 and 720)

Answer 1

did your teacher or textbook learn you any tricks to solve these kind of problems?

Answer 2

yes

Answer 3

alright, does it involve prime factorization?

Answer 4

great

Answer 5

now, since you only need to know the number of divisors and not the divisors itself, you can simply add 1 to each of the powers (so 4+1, 2+1 and 1+1 (there is an imaginary 1 above the 5)) and multiply the resulting numbers together.

Answer 6

30

Answer 7

so, 4+1 * 2+1 * 1+1

Answer 8

indeed

Answer 9

does 30 divisors involve 1 and 720

Answer 10

yes

Answer 11

do you want me to explain why this trick works?

Answer 12

yes

Answer 13

ok, so you already divided the number into the prime factors, which are of course the smallest parts that make up the number, sort of like atoms.

Answer 14

but obviously the number will also be able to be divided by any multiplication of those numbers, up and including the number itself (720).

Answer 15

so for example 5^0 * 3^0 * 2^0 = 1 will be able to divide 720 and so does 5^0 * 3^2 * 2^3

as long as you use the prime numbers inside the number in any amount that does not exceed the prime factors you will be able to divide the number by it.

So all we just did was calculate the number of possible combinations that apply to that rule

Answer 16

so for example 5^2 * 3^1 *2^1 would not work because,. even though it uses the same numbers, you are using too many 5's (as the prime factors only use a single 5, not 2)

Answer 17

ok

Answer 18

if you have any more questions feel free to ask

Answer 19

u mean the power of 2,3 and 5 shouldnt exceed 4,2 and 1

Answer 20

exactly

Answer 21

because otherwise you would exceed the number 720 or get numbers that do not divide 720.

Answer 22

but any combination inside those limits works, including to the power of 0 (the 0 is why you add an extra +1 to each power when you calculate the combinations, otherwise you are not counting the 0)

Answer 23

so in the end you get 2^0 OR 2^1 OR 2^2 OR 2^3 OR 2^4 multiplied with 3^0 OR 3^1 OR 3^2 multiplied with 5^0 OR 5^1. The number of combinations possible for that is of course 30, as you calculated before. Because there 5 'options' for the 2's and 3 'options' for the 3s and 2 'options' for the 5.

Answer 24

ok

Answer 25

thns