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# how would you go about finding out the following: how many

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how would you go about finding out the following: how many [#permalink]  25 Jul 2008, 14:52
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how would you go about finding out the following:
how many factors, that are not multiples of 6, exist in 264,600?
pls show work/explanation.
thx
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Re: PS factors [#permalink]  25 Jul 2008, 16:00
young_gun wrote:
how would you go about finding out the following:
how many factors, that are not multiples of 6, exist in 264,600?
pls show work/explanation.
thx

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

Any combination of multiples based on these prime factors that do not include both 2 and 3 are factors of 264,600 that aren't divisible by 6.
Let w = 2^n where n is any integer such that 0 <= n <= 3
Let x = 3^n where n is any integer such that 0 <= n <= 3
Let y = 5^n where n is any integer such that 0 <= n <= 2
Let z = 7^n where n is any integer such that 0 <= n <= 2

A factor of 264,600 is not a multiple of 6 if it is w*y*z or x*y*z

Number of multiples divisible by 2 but not 6 = 4*3*3
Number of multiples divisible by 3 but not 6 = 4*3*3

Total number of factors not divisible by 6 = 2*4*3*3 = 72
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Re: PS factors [#permalink]  25 Jul 2008, 16:43
Prime factorize 264,600:

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

We know that any number will be a factor of 264,600 as long as we can write it as:

2^a * 3^b * 5^c * 7^d

and a <=3, b <= 3, c <= 2, d <= 2.

There are 4*4*3*3 = 144 factors in total (add one to each power and multiply). We only want to count factors that are not multiples of 6. Let's count the factors that are multipes of 6: then a can only be 1, 2 or 3, and b can only be 1, 2 or 3. c and d can each still be 0, 1 or 2. We have 3*3*3*3 factors, or 81 factors, that are multiples of 6. We should then have 144 - 81 = 63 factors that are not multiples of 6.
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Re: PS factors [#permalink]  25 Jul 2008, 16:49
zoinnk wrote:
young_gun wrote:
Number of multiples divisible by 2 but not 6 = 4*3*3
Number of multiples divisible by 3 but not 6 = 4*3*3

Nice solution, but you've double-counted the case where the exponent on both two and three is zero (that's included in both cases that you consider), which is why your answer is nine larger than the correct answer. That is, there are nine factors which are multiples of neither two nor three- we only want to count these once, not twice.

In fact, the number of multiples of 2 which are not multiples of 6 is 3*1*3*3 = 27 (the power on 2 can be 1, 2 or 3, and the power on 3 must be 0), not 4*3*3. The same is true of multiples of 3 which are not multiples of 6.
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Re: PS factors [#permalink]  26 Jul 2008, 09:42
IanStewart wrote:
zoinnk wrote:
young_gun wrote:
Number of multiples divisible by 2 but not 6 = 4*3*3
Number of multiples divisible by 3 but not 6 = 4*3*3

Nice solution, but you've double-counted the case where the exponent on both two and three is zero (that's included in both cases that you consider), which is why your answer is nine larger than the correct answer. That is, there are nine factors which are multiples of neither two nor three- we only want to count these once, not twice.

In fact, the number of multiples of 2 which are not multiples of 6 is 3*1*3*3 = 27 (the power on 2 can be 1, 2 or 3, and the power on 3 must be 0), not 4*3*3. The same is true of multiples of 3 which are not multiples of 6.

Yep, you're right I knew I was going to miss something dumb.
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Re: PS factors [#permalink]  28 Jul 2008, 17:27
zoinnk wrote:

Yep, you're right I knew I was going to miss something dumb.

Actually i think that you are right. and Ian's post was a response to young_gun.

2^3 3^3 5^2 7^2 is the simplified expression.

Now going by zoinnk's method, total number of factors = 144.
since power of 2 and 3 cannot be 0 at the same time, total number of factors that are multiples of six is indeed 3*3*3*3=81. 144-81 is the right answer.

Going by the other direct approach where you compute non multiples of six.. you have

case 1) power of 2 is 0. So power of 3 can be 0,1,2,3. Hence number of desired factors = 4x1x3x3 = 36

case 2) power of 3 is 0. Power of 2 can be 1,2,3 and not zero. 0 means only having 5^2 and 7^2 .. but this has already been taken in case 1.
so here we have 1*3*3*3 = 27

total = 36+27 = 63
Re: PS factors   [#permalink] 28 Jul 2008, 17:27
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