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Old 07-16-2008, 05:27 PM   #26
Brad Pittiful
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the answer is 42
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Old 07-16-2008, 05:30 PM   #27
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Ok, I'm going with 2:
45 -> 15, 9, 5, 3
75 -> 25, 15, 5, 3
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Old 07-17-2008, 12:19 AM   #28
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Quote:
Originally Posted by elucas730 View Post
Ok, I'm going with 2:
45 -> 15, 9, 5, 3
75 -> 25, 15, 5, 3
That can't be right because 45 and 1 are factors of 45.
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Old 07-17-2008, 12:21 AM   #29
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SOLUTION SPOILER

Ok, I started thinking about the problem if you add the condition that it has to be distinct factors and you allow 1 to be a factor.That problem could then be restated as follows: How many integers less than 100 can be expressed as the product of two odd prime numbers. Since the number itself and 1 are ALWAYS going to be factors and we can't have more than two prime factors then we just need the other two factors to be prime.

I'm still at a loss at how to do the problem that I posed though.

Last edited by Physics Junkie; 07-17-2008 at 04:11 AM.
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Old 07-17-2008, 12:28 AM   #30
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Quote:
Originally Posted by mprziv
You're kidding?

Um, that would make a load of answers.

1*1*1*1
1*1*1*3
1*1*1*5
1*1*1*7
1*1*1*9
MPRZIV, 1*1*1*9 wouldn't work because 9 there are not 4 factors--only 3 in that case: 1,3,9. 1*1*1*7 is the same deal--only 2 factors. This problem is hell.

Last edited by Physics Junkie; 07-17-2008 at 03:18 AM.
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Old 07-17-2008, 03:14 AM   #31
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wat??
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Old 07-17-2008, 03:44 AM   #32
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SOLUTION SPOILER

Quote:
Ok, I started thinking about the problem if you add the condition that it has to be distinct factors and you allow 1 to be a factor.That problem could then be restated as follows: How many integers less than 100 can be expressed as the product of two odd prime numbers. Since the number itself and 1 are ALWAYS going to be factors and we can't have more than two prime factors then we just need the other two factors to be prime.

I'm still at a loss at how to do the problem that I posed though.
Ok. The same principle applies but I forgot a part that I'll get into below. I started thinking about how you can count the factors of a number. First you get the prime factorization. Then the total number of factors is equal to the exponents of the prime numbers+1 all multiplied by eachother. So for 125 = 5^3 it has (3+1) factors: 1, 5, 25, 125. For 72 you have 2^3 * 3^2 so you have exactly (3+1)*(2+1) factors giving you 12 factors in total: 1, 2, 3, 4, 6, 8, 9, 12, 24, 27, 36, 72,

This is essentially counting the combinations of different factors you could make from its prime factorization. You add 1 because you need to count the fact that you can raise any exponent to the power of 0.

So now we know that we are looking for numbers from 1 to 99 that are the product of two distinct prime numbers that are ODD. So we can eliminate 2 even though it is prime(pardon the pun). 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, and we stop there because the next prime is 37 and 3*37 > 100.

So now our question is how many unique pairs of prime numbers can we pick from this list that when multiplied are less than 100. Here is where it becomes a turn the crank problem. NINE can be made with the number 3, FIVE can be made with the number 5(excluding 3*5 since we already counted that with 3), and TWO can be made with the number 7. So our result is 9+5+2 = 16 numbers so far. But we forgot(read below).

15, 21, 33, 39, 51, 57, 69, 87, 93
35, 55, 65, 85, 95
77, 91

So yay all of our answers so far are unique.

But we forgot one thing. What about prime*(prime^2) or prime*(prime^3)?

Now it's more of a "turn the crank" problem than before.

3*(3^2) = 27. The factors are 1, 3, 9, 27

That's the one we left out. We can't do prime*another_prime^2 because then we have more than 4 factors.

****. This method sucks. There has to be some way to do this easily in math beyond my understanding.

So the question doesn't ask for distinct factors and that's so that we can include the last #%*&@&!!&%# of an element: 27.

Last edited by Physics Junkie; 07-17-2008 at 04:12 AM.
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Old 07-17-2008, 03:56 AM   #33
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Eleventy.
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Old 07-17-2008, 04:09 AM   #34
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MPRZIV/Capt Crunch, is there a better way than this? Something to do with moduli counting perhaps?
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Old 07-17-2008, 04:15 AM   #35
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Quote:
Originally Posted by dmross View Post
Am I right?
You are right about the format...almost. I didn't see where you were coming from the first time I read this but now I see that you are on the right track but got the wrong answer and forgot a case. Quick thinking on that. What's your background?
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Old 07-17-2008, 10:38 AM   #36
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Quote:
Originally Posted by Physics Junkie View Post
You are right about the format...almost. I didn't see where you were coming from the first time I read this but now I see that you are on the right track but got the wrong answer and forgot a case. Quick thinking on that. What's your background?
Yeah I missed the prime*prime^2 and prime*prime^3 possibility for 27. But why don't the numbers 6, 8 and 10 count with the original wording in the subject line?

1 2 3 6 - four distinct factors
1 2 4 8 - four distinct factors
1 2 5 10 - four distinct factors

As for my background, I have a BS in CS.
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Old 10-06-2009, 09:30 PM   #37
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So I'm studying for the GMAT right now and came across this thread doing a search for "GMAT". Even though no one here probably cares anymore, I just thought I'd throw it out there that the GMAC uses the word distinct differently than many people here are using it. The GMAC uses it in the context of "How many distinct prime factors does 36 have?" The answer is 2 (2 and 3). If the question was asked about the number 72, the answer would also be the exact same. I seriously doubt any GMAT or SAT question would ask the original question in the title without specifying that they are looking for distinct prime factors, not just distinct factors. It would be way too cumbersome. Assuming that's the case, the least possible value of a number with 4 distinct prime factors is a number divisible by 2,3,5, and 7, which is 210. So the answer is 0 numbers less than 100 have 4 distinct prime factors.
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