Manhattan Challenge Problem

Any approach to this one...........?

I would think you begin by breaking it into its prime factors and finding how many possible combinations of those primes fit the category. I dont have an answer but hopefully someone can shed some light on it.

I found all factors of 264600 and subtracted from them the factors of 6....I got 63

800GMAT wrote:How many numbers that are not divisible by 6 divide evenly into 264,600?

(A) 9
(B) 36
(C) 51
(D) 63
(E) 72

ok, long shot but here goes...

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

The numbers that are not divisible by 6 should either not have a 3 or not
have a 2 in it as the product would yield a multiple of 6.

Combinations are 5^x*7^y*3^z and 5^x*7^y*2^z

Combinations for 5^2 * 7^2 * 3^3 = 3 * 3 * 4 = 36
(i.e. take 5^0, 5^1, 5^2 - 3 factors etc.)

Similarly, combinations for 5^2 * 7^2 * 2^3 = 3 * 3 * 4 = 36

Adding up, we get 72.

Out of this, we have to prune duplicates as 2^0 = 3^0 = 1. Total of
3*3 (possible powers of 5 * possible powers of 7) = 9

So, total = 72 - 9 = 63

If this answer is wrong, forget all of the above

Here is the link to the OA and the OE
https://www.manhattangmat.com/ChallProbLastWk.cfm

If such problems appear on GMAT, I will stop preparing for it.

800GMAT wrote:Here is the link to the OA and the OE
https://www.manhattangmat.com/ChallProbLastWk.cfm

Phew .

I will go with D

Yeah I will roll the dice on this one.....gone are the days when I used to delve into a problem and spend sleepless nights

So the primes of 264600 are 2, 2, 2, 3, 3, 3, 5, 5, 7, 7

combinations that don't make 6 are (a) the 2s 5s 7s (b) 3s 5s 7s

(a) 2^3 * 5^2 * 7^2 = 36 combinations
(b) 3^3 * 5^2 * 7^2 = 36 combinations

so total is 72. This rules out A and B

But there are duplicates e.g. the combinations of 5^2 and 7^2 but I don't know how to remove them. So i know the answer is less than 72 and definitely not 9 or 36. So it is between 51 and 63... Intuituvely, 51 seems a little low coz that'd mean there were 21 duplicates so I'd guess the answer to be 63 but I am not sure.

You are right this is a tough one to do in 2 mins coz you have to prime factorize and then do combinations excluding 6 and then remove duplicates = three layers... definitely a question I think I'd solve up till I got to 72 but then I'd be stuck and have to guess...

Solution in simple words:
if we factorize a number in prime factors in the form p^m*q^n*r^l..
the total number of factors this number can have can be given by (m+1)(n+1)(l+1)...

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

now the divisor should not be divisible by 6. It is possible only when the divisor is NOT a multiple of 2 AND 3.
so all number with 2^3,5^2 and 7^2 AND all numbers formed with 3^3, 5^2 and 7^2 are the candidates.
Case 1: all number with 2^3,5^2: 4*3*3=36
case 2: 4*3*3=36
total 72.

point to consider
while counting the numbers, we counted the numbers formed with only 5^2 and 7^2 in both cases. total numbers possible with 5^2 and 7^2=3*3=9.
we need to subtract it from the result: 72-9= 63 answer

another apporach.

total no of possible factors= 144.
total number that has 6 as common factor= 81
so total number of factor that doesn't have 6= 144-81 = 63

The question is asking about the factors of 264,600

We know from the divisibility rules that the factors are : 5^2X7^2X3^3X2^3

we want to find the factors which are not divisible by 6

There are 36 factors with out 2 , 12 with 2 and 5 , 12 with 2 and 7
and 3 factors of 2 (2,4and 8) so totally = 36+12+12+3 =63

I'm sorry I may have missed something here, but how do you get the combinations that equal 36?

I see 3*3*4 but how do you arrive at 3, 3 and 4???