Problem Solving

goyalsau wrote:

sanju09 wrote:

goyalsau wrote:what to do in this problem?

264600 = 2^3 × 3^3 × 5^2 × 7^2, in this, not divisible by 6 are contained in 2^3 × 5^2 × 7^2 and 3^3 × 5^2 × 7^2 only.

2^3 × 5^2 × 7^2 has a total of 4 × 3 × 3 = 36 such divisors and 3^3 × 5^2 × 7^2 has a total of 4 × 3 × 3 = 36 such divisors.

In all those 36 + 36 = 72 such divisors, the divisors generated by 5^2 × 7^2, which are 3 × 3 = 9 in number, are included twice; hence we need to subtract 9 from 72 to answer [spoiler]63 as the most appropriate one.

D[/spoiler]

Sanju i was trying to solving differently but I am doing some mistake but not able to get the answer.
Brother if can let me know my mistake i will be really helpfull

264600 = 2^3 × 3^3 × 5^2 × 7^2
IN all 4 * 4 * 3 * 3 = 144 factors

Now we want 6 out
then 2^3 * 3^3 * 5^2 = 4 * 4* 3 = 48 factors
and 2^3 * 3^3 * 7^2 = 4 * 4* 3 = 48 factors

in all 96 factors
common factors 2^3 * 3^3 = 4 * 4 = 16
48 + 48 = 96 - 16 = 80

Now we subtract 144-80 = 64
but the answer is 63 so i want to know which one factor i am missing

I know its hard to find problems than doing it all along but i will really thank full if you can let me know my mistake.

You are not trying to answer the central question here, because you are not trying to get the divisors NOT divisible by 6, this way. You would not have included "2^3 X 3^3" to factor considerations if you really wanted 6 OUT.

Address a different problem, get a different result, no marvel!

264,600 = 2^3 * 3^3 * 5^2 * 7^2

Any combination of these prime factors will yield a factor of 264,600.

For each prime factor, the number of choices = exponent + 1.

If the exponent is 2, we'll have 3 choices for that prime factor. This is because in our combination we can use none of that prime factor, 1 of that prime factor, or 2 of that prime factor, giving us 2+1=3 choices for that prime factor.

If the exponent is 3, we'll have 4 choices for that prime factor. This is because in our combination we can use none of that prime factor, 1 of that prime factor, 2 of that prime factor, or 3 of that prime factor, giving us 3+1=4 choices for that prime factor.

We need to count all the prime-factor combinations that are not divisible by 6. Since 6=2*3, a good combination may include 2 or 3 but not both:

2^3 * 5^2 * 7^2 gives us 4*3*3 = 36 factors
3^3 * 5^2 * 7^2 gives us 4*3*3 = 36 factors

In the combinations above, we've double-counted the combinations that include only 5 or 7:
5^2 * 7^2 gives us 3*3 = 9 factors that have been counted twice.
We need to subtract these 9 factors from our total.

36+36-9 = 63.

The correct answer is D.

goyalsau wrote:

sanju09 wrote:

goyalsau wrote:what to do in this problem?

264600 = 2^3 × 3^3 × 5^2 × 7^2, in this, not divisible by 6 are contained in 2^3 × 5^2 × 7^2 and 3^3 × 5^2 × 7^2 only.

2^3 × 5^2 × 7^2 has a total of 4 × 3 × 3 = 36 such divisors and 3^3 × 5^2 × 7^2 has a total of 4 × 3 × 3 = 36 such divisors.

In all those 36 + 36 = 72 such divisors, the divisors generated by 5^2 × 7^2, which are 3 × 3 = 9 in number, are included twice; hence we need to subtract 9 from 72 to answer [spoiler]63 as the most appropriate one.

D[/spoiler]

Great Work Sir
I was always sure that you are the one...
Thanks.

Hi goyalsau, if it is really so then how did you find GMATGuruNY explanation different from sanju09?