Assistance needed

Perhaps a "mathematical way" would be to consider the prime factorization, [itex]180 = 2^2 \cdot 3^2 \cdot 5 [/itex]. Therefore any factor of 180 which is a multiple of 6 is of the form [itex] 6 \cdot 2^a \cdot 3^b \cdot 5^c [/itex] where [itex] 0 \le a,b,c \le 1 [/itex]. Now it is easy to count how many possibilities there are.

 

We know that the prime factors of 180 = 2*2*3*3*5.
So, we can take (2*3) *2 = 6*2 =12 which is a multiple of 6 and a factor of 180.
Next, take the prime factors 2*3 * 3 = 6*3 = 18. Thus, it is a multiple of 6 and a factor of 18.
Continue, you see a pattern i.e. (2*3)*5, (2*3)*3*3... of permutations.

 

Hi Petkovsky!

Hint: if 6n is a factor of 180,

that means that 180/6n is a whole number,

and so … ?

 

Physicsissuef, don't give the answer!

(delete it if you still can)

(i'm not sure it's right anyway …)

Just give hints.

 

6*2*3*5

Yes, my answer wasn't correct actually :D

6*2, 6*3, 6*5

Maybe this is correct.

 

Physicsissuef, what part of "don't give the answer!" was not clear?

You know we're only supposed to give help, not complete answers!

Well … you're obviously just guessing now … so no harm done!