Answer requires justification:
98000 = 2^4 * 5^3 * 7^2.
(a) Compute the number of distinct divisors of 98000
(b) Compute the number of odd divisors of 98000
Anyone got any ideas how to do this
Answer requires justification:
98000 = 2^4 * 5^3 * 7^2.
(a) Compute the number of distinct divisors of 98000
(b) Compute the number of odd divisors of 98000
Anyone got any ideas how to do this
Often times when I'm stuck on a math problem, I find it immensely helpful to consider a smaller/simpler version of the problem, to help me get a feel for what's going on. Then I can attempt to scale my methods up to the real problem and see if it works the same (it usually does).
So consider how you might tackle the problem if asked about 18 = 2 * 3^{2}. What are all of the factors of 18? If you factorize them into primes, what do you notice? Also try writing them but including the zero powers (i.e. 3 = 2^{0} * 3). Does that change what you noticed? Now maybe try the problem with 108 = 2^{2} * 3^{3}. Repeat the same process, writing down all of the factors of 108 and writing as products of primes. What do you notice? Does this fit with what you noticed about 18? Finally, do the same for 360 = 2^{3} * 3^{2} * 5.
Use what you learned in pre-algebra about prime factors, and other divisors. In how many ways can powers of 2 be factors? In how many ways can powers of 5? of 7?
In how many ways can 2*(a power of 5) be factors? 2*(a power of 7)? And so forth. Consider the various cases.
For part (b), find the divisors that do not include any 2's.
Hope this is not giving too much away
When I was trying to figure out what stapel was saying, I realized it was useful to write the product of prime factors as
(2*2*2*2)*(5*5*5)*(7*7)
and then to realize that any unique combination of factors (in the permutations and combinations sense) that you can make out of this set is a divisor of the whole thing.
Because, in a sense, when you do the division, you're choosing a subset of the numbers in the product above to "cross out". How many unique subsets are there? Don't forget that this product has also been multiplied by a factor of 1, implicitly (you can string an arbitrary number of 1s onto the above product w/o changing the answer).
Hope that helps
Last edited by j-astron; Today at 10:28 AM. Reason: Streamline wording, prevent confusion
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