How many numbers?

[aras82]

How many numbers that are not divisible by 6 divide evenly into 264,600?

[touche]

2^3.3^3.5^2.7^2
all multiples of these leaving aside when 2 are 3 are together is what we are looking for. Lost here,,....
How do we find how many from here?

[EdYo]

I've seen a similar problem in my math club, 4 years ago, however I didn't remember the faster way to solve this problem, so i used whatever i can remember.

264600 = 2^3*3^3*5^2*7^2

I thought about the factors so i googled the method to calculate the total factors of a number (here is the link I found Math Forum - Ask Dr. Math)

If you factor a number into its prime power factors, then the total
number of factors is found by adding one to all the exponents and
multiplying those results together. Example: 108 = 2^2*3^3, so the
total number of factors is (2+1)*(3+1) = 3*4 = 12.

This looks similar to the Venn diagram problem to me, so i use Venn diagram to solve it. Can't draw it here and I'm too lazy to scan.

P(A and B) = P(not A) + P(not B) - P(A nor B)

P(not A) = number of factors is not multiple of 2 --> should include : (7^2)(3^3)(5^2) --> apply the formula i googled --> P(not A) = (2+1)*(3+1)*(2+1)= 36

P(not B) = number of factors is not multiple of 3 --> should include : (7^2)(2^3)(5^2) --> P(not B) = (2+1)*(3+1)*(2+1)= 36

P(not A nor B) = number of factors is not multiple of 2 or 3 --> should include : (7^2)(5^2) --> P(not A nor B) = (2+1)*(2+1)= 9

==> P(A and B) = 36 +36 - 9 = 63

What is the OA ?

[3Gs]

I guess it is (4*4*3*3) - (multiplication of power of 2 & 3)
= 144 - (3*3)
= 135

I AM NOT SURE

[Fiver]

In order that it is not divisible by 6, the number should have either 3 or 2 as one of it's factors, but not both.
At the same time it must be a multiple of 264,600 for it to evenly divide into 264,600.
I find none, as for a number to be a multiple of 264,600 it must have atleast 3 2s and 3 3s.
I wonder if i have correctly interpreted the question.

[finsisher]

144-81= 63.. 63 is the correct answer..
I guess every one knows how to get 144 for 81 ... since we need atleast one 2 and one 3 so for two and three powers correspondingto 2 = 3 and power corresponding to 3 =3 we have to ignore 2^0 and 3^0... hence their powers are not increased by 1

[3Gs]

Hi finsisher...How did you calculate 81.. could you please elaborate...? Thanks

[Fiver]

Ahah! Now I understand the meaning of this question.
Its asking for the no. of even factors of 2464600 that are not divisible by 6.
264600 = 2^3*3^3*5^2*7^2
Total factors = 4*4*3*3 = 144
Total factors divisible by 6 = 3*3*3*3=81
As finisher suggests we need to ignore 2^0 and 3^0.
Hence the answer is 63.