Tuesday

January 17, 2017
Posted by **Anonymous...please help!** on Monday, March 11, 2013 at 3:01am.

- geometry -
**Lena**, Monday, March 11, 2013 at 3:40amIs there a figure that you were given? This question does not make much sense to me.

- geometry -
**Anonymous...please help!**, Monday, March 11, 2013 at 4:11amno.....there was no figure

- geometry -
**Lena**, Monday, March 11, 2013 at 4:35amWhat exactly is the question asking?

It would make sense if they were asking the greatest distance between 2 points but asking what shortest distance between 2 points does not make sense.

Perhaps I'm not understanding the question properly. If you could expand on what the question is asking perhaps I could show you how to do the math involved. - geometry -
**Anonymous...please help!,**, Monday, March 11, 2013 at 4:39ami have another problem can u try it???

- geometry -
**Lena**, Monday, March 11, 2013 at 4:44amSure; post it and I'll give it a go.

- geometry -
**Anonymous...please help!,**, Monday, March 11, 2013 at 4:46amok

- geometry -
**Anonymous...please help!,**, Monday, March 11, 2013 at 4:48amHow many numbers from 1 to 100 are multiples of 3 but not 5?

- geometry -
**Lena**, Monday, March 11, 2013 at 5:01amOK.

So there are 33 multiples of 3 that are < or = 100.

3,6,9,12,15,18,21,24,27,30,33, 36,39,42,45,48,51,54,57,60,63,66,69,72,75,78,81,84,87,90,93,96,99 ect.

There are 20 multiples of 5 that are < or = 100.

5,10,15,20,25,30,35,40,45,50,55,60,65,70,75,80,85,90,95,100

The ones that match are the one's you would cancel out: 15,30,45,60,75,90

33-6 = 27

NOW THE EASIER WAY TO DO THIS:

I think the easier way to do this would be to figure out the # of multiples of 3 that are < = 100 which is 33 and then subtract all of the multiples of 15 (3x5) up to 100 and there are 6 (15,30,45,60,75,90)

33-6 = 27 - geometry -
**Steve**, Monday, March 11, 2013 at 10:45amNot sure how to do the hexagon problem, but it's asking how to figure the greatest minimum separation of points in the hexagon.

If you try placing them at the vertices and the center, then the smallest distance is the side length. Surely they can be distributed in such a way that the two closest points are farther apart than that.

Gotta think on it some. Maybe work my way up from fewer points or a different polygon. - geometry -
**Steve**, Monday, March 11, 2013 at 11:30amActually, it appears that the center and vertices are the optimal positions, since the radius of the circumcircle is the same as the side length.

If you move any point on the vertices, it will be closer to the center. If you move the point on the center, it will be closer to one of the vertices.

I'm sure this happy arrangement only works for the hexagon, where the radius is equal to the side length. - geometry -
**adil**, Tuesday, March 12, 2013 at 1:32amsteve-ur right.. we can place six points on each of it's veritices...nd if we put the other point on it's center,the maximum distance between the two closest points will be same as the length of it's side length...so the answer is 20.