I know how to list all the factors, but please solve this using prime factorization.
1200=2^4*3*5^2, which means it has 30 factors(5*2*3), but what now?
how many even and odd positive factors does it have?
Please feel free to leave a measage about any thoughts you have about this question.
1200=2^4*3*5^2, which means it has 30 factors(5*2*3), but what now?
how many even and odd positive factors does it have?
Please feel free to leave a measage about any thoughts you have about this question.
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There's a cool AIME problem just like this! I think it's AIME 2000 or 2001. Look it up on the Art of Problem Solving. (Great website!)
You can do a few things. The easiest thing to do is to count the odd factors. You're right that there are 30 factors total. (I'm assuming you did the math right. I can't factor 1200 in my head this late at night.) Which ones are odd, though?
They're the ones that DON'T have a 2 in them! Normally you're allowed to choose 0, 1, 2, 3, or 4 factors of '2', 0 or 1 factors of '3', and 0, 1, or 2 factors of '5'.
Let's make sure that they're odd, by SETTING the number of factors of 2 equal to ZERO! That limits the number of choices to 2 choices for '3', and 3 choices for '5'. So we get 2 * 3 = 6 odd factors.
Thus, there are 30 - 6 = 24 even factors.
You can also find the number of even factors by recognising that an even factor must have at LEAST one factor of 2. So you have only FOUR choices for 2^x, while you still have 2 and 3 choices for your 3 and 5 exponent. Thus you have 4 * 2 * 3 = 24 even factors.
Or, you can do it by ratios. Let's say x is an odd factor of 1200. Then 2x, 4x, 8x, and 16x are even factors of 1200. Thus for every one odd factor, there are four even factors. Thus, the factors come in groups of 5. Since there are 30 factors total, there are six of these groups of five. Thus there are six odd factors, and 6*4 = 24 even factors.
Followup questions: How many positive factors of 1200 don't end in zero? How many positive factors of 1200 are composite? How many factors of 1200 are neither prime nor composite?
You can do a few things. The easiest thing to do is to count the odd factors. You're right that there are 30 factors total. (I'm assuming you did the math right. I can't factor 1200 in my head this late at night.) Which ones are odd, though?
They're the ones that DON'T have a 2 in them! Normally you're allowed to choose 0, 1, 2, 3, or 4 factors of '2', 0 or 1 factors of '3', and 0, 1, or 2 factors of '5'.
Let's make sure that they're odd, by SETTING the number of factors of 2 equal to ZERO! That limits the number of choices to 2 choices for '3', and 3 choices for '5'. So we get 2 * 3 = 6 odd factors.
Thus, there are 30 - 6 = 24 even factors.
You can also find the number of even factors by recognising that an even factor must have at LEAST one factor of 2. So you have only FOUR choices for 2^x, while you still have 2 and 3 choices for your 3 and 5 exponent. Thus you have 4 * 2 * 3 = 24 even factors.
Or, you can do it by ratios. Let's say x is an odd factor of 1200. Then 2x, 4x, 8x, and 16x are even factors of 1200. Thus for every one odd factor, there are four even factors. Thus, the factors come in groups of 5. Since there are 30 factors total, there are six of these groups of five. Thus there are six odd factors, and 6*4 = 24 even factors.
Followup questions: How many positive factors of 1200 don't end in zero? How many positive factors of 1200 are composite? How many factors of 1200 are neither prime nor composite?