Question

Find the sum of all positive rational numbers that are less than 5, and that have a denominator of 30 when written in lowest terms.

Answer 1

I have no idea?

Answer 2

Hmmm... well, to be a positive, rational number that is less than 5 and has a den'r of 30, the number has to be of the form \[\frac{ a }{ 30 }\]where some very specific limitations can be placed on a. Can you think of what they are?

Answer 3

they have to have no common factors?

Answer 4

Hmm.... wait. This is more complicated than I initially thought, lol. But yes, they have to have no common factors... so the numerator can't be a multiple of 2, 3, or 5.

Answer 5

We also know a lower and upper bound on a.... right?

Answer 6

0-5?

Answer 7

no.

first of all for lower bound, we are looking for positive number so must have a>0, so a=1 is the smallest possible a.

Answer 8

And since we want a/30<5, we need to have a<.... what?

Answer 9

150

Answer 10

Right... so 0<a<150, and no factors of 2,3 or 5.

Now I'm kind of stuck. Sorry, I thought I had an idea where to go, but I'm not sure what to do next. It actually wouldn't be that hard to cook up the list by hand, but I doubt that such "brute force" is what's expected. There must be some slicker method.

Answer 11

Wait, I have an idea.... remember, what we need is a way to sum up these terms in the numerator. Once we have that sum, they are all over the LCD of 30. So if we take\[\large \sum_{n=1}^{149}n\]that gives ALL the numbers in that interval. the problem with that, of course, is that it includes all the numbers that have factors in common with 30. But what if we do:\[\large \sum_{n=1}^{149}n-\sum_{n=1}^{74}2n-\sum_{n=1}^{49}3n-\sum_{n=1}^{29}5n\]Do you see what that does?

Now if I could just remember how to evaluate those sums... lol.

Answer 12

OK, I think the sums are n(n+1)/2, but we have to pull out the constants.

Annnnd... I'm talking to myself so I'm outta here. Respond and/or tag me if you want further help.

Answer 13

Ughh...I'm so rude. I had to get off because of an...emergency. Sorry!

Answer 14

lol 'sok, thanks for coming back to explain. You are allowed to have a life. :)

Answer 15

First, let's look at all such rational numbers between
0 and 1.

Answer 16

100

Answer 17

What?? @jonjenkins7653 the rational numbers are all supposed to be less than 5. Did you read my discussion above?

Answer 18

Aww...thanks for understanding debbie

Answer 19

I think it works.\[\large \sum_{n=1}^{149}n-\sum_{n=1}^{74}2n-\sum_{n=1}^{49}3n-\sum_{n=1}^{29}5n\]\[=\large \sum_{n=1}^{149}n-2\sum_{n=1}^{74}n-3\sum_{n=1}^{49}n-5\sum_{n=1}^{29}n\]\[=\frac{ 149(150) }{2 }-2\frac{ 74(75) }{2 }-3\frac{ 49(50) }{2 }-5\frac{ 29(30) }{2 }\]

Answer 20

@DebbieG Now you have to add back on the multiples of 2 and 3 (i.e. 6) because you have taken them away twice. Similarly with the multiples of 10 and 15. And finally take away the multiples of 30, since you took those away three times but added them back on three times!

Answer 21

I think?? That would give the numerator.....

Oh wow! Yes of course, I see what you mean @Erin001001 !

Answer 22

this is a great question!

Answer 23

wait...what? I'm so confused

Answer 24

sum all the primes from 7 to 150 and divide by 30.

any other number will be a multiple of 2, 3, 5 or some combination of thos numbers.

Answer 25

of course, 150 isn't prime but i hope you get the idea...

Answer 26

I was originally thinking of just doing something with primes. Wasn't sure if that was the way to go though.

Answer 27

the numerator must be relatively prime to 2, 3 & 5... so all the primes and multiples that don't include those.
so sorry, not just the primes...

Answer 28

That's basically what I was getting at above, when I said doing it by "brute force". lol That would work, since the number of terms in the sum is reasonably small. But it seems like there should be a more algorithmic way to approach it, lol. :)

Answer 29

the sum of the terms is 2266

Answer 30

Right @pgpilot326 ... I think if done by brute force, you just have to take all integers from 1 to 149 that aren't a multiple of 2,3 or 5. So a number like 77 is not prime, but IS part of the sum.

Answer 31

@orple8 2266? Is that 2266/30? Intuitively, 2266 seems too big to me.

Answer 32

http://www.wolframalpha.com/input/?i=sum+all+the+primes+from+7+to+150

Answer 33

yeah, i jumped to soon...
it just seems that the including everything then adding and subtracting different multiples is a bit confusing.