
You should always read "Find the last two digits" as "calculate stuff mod 100".
The described method works, because the order of Z/100Z (as a ring) is phi(100). This means that the order of any element of Z/100Z is a factor of 40 and thus, g^40 = 1 for any member of Z/100Z.
You could also simply try to find the order of 3 (mod 100) by calculating 3 9 27 81=19 57=43 121=21 63 189=89=11 33 99=1
and thus derive that the order of 3 mod 100 is actually 10, and that you can thus remove any multiples of 10 from the exponent in question, once again leading you to 3^3 as your result. Takes a bit longer, but not that much longer, and takes less stuff you need to realize. This is probably how i would have done it.
Didn't we have a similar question a few pages back?

United States23581 Posts
On May 18 2019 02:19 Simberto wrote: You should always read "Find the last two digits" as "calculate stuff mod 100".
The described method works, because the order of Z/100Z (as a ring) is phi(100). This means that the order of any element of Z/100Z is a factor of 40 and thus, g^40 = 1 for any member of Z/100Z.
You could also simply try to find the order of 3 (mod 100) by calculating 3 9 27 81=19 57=43 121=21 63 189=89=11 33 99=1
and thus derive that the order of 3 mod 100 is actually 10, and that you can thus remove any multiples of 10 from the exponent in question, once again leading you to 3^3 as your result. Takes a bit longer, but not that much longer, and takes less stuff you need to realize. This is probably how i would have done it.
Didn't we have a similar question a few pages back?
I am sure we did. I appreciate everyone's patience with me in this thread, since it's mostly me asking questions here I realize I am just absorbing everyone's knowledge and don't really have an opportunity to contribute much. Maybe now that I am back on ADD medication I'll be able to learn this stuff a bit quicker.
ok so, This means that the order of any element of Z/100Z is a factor of 40 and thus, g^40 = 1 for any member of Z/100Z seems to be the crucial part. I am sure this will be on the exam, thank you.
I actually like your method too, though. I remember that being done in this thread, and it's pretty straightforward if I can remember to use the negative modulus

On May 18 2019 02:09 travis wrote:Here is another question. I don't mean to spam, I just think it's better in a separate post. We have a 3d cube, each point(vertex) is labeled 1...8 G is the group of rotational symmetries of this cube Is G cyclic? Give a brief justification I say: Show nested quote + No, it is not. Any of these permutations are their own inverse in the group and thus the group is not cyclic.
Is my answer accurate? Would i get full credit for that?
Your answer is correct, but if you would get full credit relies on how explicitly you have established some of those things in class.
I would go a bit more into detail here, and basically find two different objects of order 2 (180° Rotation on x and y axis, for example), and then explain why there can not be more than one object with an order of 2 in a cyclical group. Thus, you have an actual proof.
I actually like your method too, though. I remember that being done in this thread, and it's pretty straightforward if I can remember to use the negative modulus
The problem with my method is that it doesn't scale well at some point. If in another situation you have an object of order 800 or so, you would have to do 800 calculations to discover this. It works well for kinda small groups, but at some point it can be really annoying, and other ways to quickly discover the result can be helpful.

On May 18 2019 02:09 travis wrote:Here is another question. I don't mean to spam, I just think it's better in a separate post. We have a 3d cube, each point(vertex) is labeled 1...8 G is the group of rotational symmetries of this cube Is G cyclic? Give a brief justification I say: Show nested quote + No, it is not. Any of these permutations are their own inverse in the group and thus the group is not cyclic.
Is my answer accurate? Would i get full credit for that?
This is true of the cyclic group Z/2Z so no. Your argument would need at least one extra detail to suffice.
EDIT: Post above this covers it fully. The extra detail could simply be observing that G>2.

United States23581 Posts
I have two finite line segments on a cartesian plane. The line segments are defined by their endpoints ((x1,y1),(x2,y2)) and ((x3,y3),(x4,y4)).
The lines are not exactly the same, but they can intersect, touch, or be completely disconnected from each other.
Here is the question:
Imagine that we form a single polygon from the endpoints of our two lines. It could be a triangle, or a trapezoid  depending on if the an endpoint of one line lays on the other line or not . Given an arbitrary point (x5,y5) on my cartesian plane, I want to know if that point lies within the bounds of the polgyon.
Hooow do I do this?

You need to find a way to describe the polygon, for example by finding some sort of "paramertrization". I think this could be done by definining a mapping f: [0,1]^2 > R^2 where f([0,1]x{0}) is the first line segment and f([0,1]x{1}) is the second line segment. Then the question is equivalent to checking if the point lays in the image of this map.

How is your original polygon even constructed? Is it just p1> p2 > p3 > p4 > p1? That would result in basically 2 triangles with a common point in the case of the two original lines intersecting. (or in cases where the connection lines intersect)
Or are you magically reordering your input points so it always ends as a single polygon? Then how is this magic reordering defined? Because there are multiple possibilities for a given input of 4 points... And if you magically reorder, then what was the meaning of those 2 initial lines anyway? "Here are two lines, they might be opposing edges or maybe diagonals of your new polygon"

United States23581 Posts
The easiest way to put it is that the constructed polgyon is the one with the maximum size, and that the original lines can be included within the bounds of the polygon (for example if the line criss crossed such that they made a shape like a plus sign or a cross, then the polygon would be the diamond around that cross)

On July 07 2019 08:41 travis wrote: The easiest way to put it is that the constructed polgyon is the one with the maximum size, and that the original lines can be included within the bounds of the polygon (for example if the line criss crossed such that they made a shape like a plus sign or a cross, then the polygon would be the diamond around that cross)
That is not guaranteeing a unique solution.

I wouldve read the assignment as taking the convex hull of the endpoints (prob mostly because convex combinations are what I usually encounter...). Then it's easy to describe the points that lie inside the set. However, I absolutely agree that the task is not worded in a welldefined way from what I can tell.

United States23581 Posts
On July 07 2019 16:48 mahrgell wrote:Show nested quote +On July 07 2019 08:41 travis wrote: The easiest way to put it is that the constructed polgyon is the one with the maximum size, and that the original lines can be included within the bounds of the polygon (for example if the line criss crossed such that they made a shape like a plus sign or a cross, then the polygon would be the diamond around that cross) That is not guaranteeing a unique solution.
Can you give me an example of why not?

On July 08 2019 04:41 travis wrote:Show nested quote +On July 07 2019 16:48 mahrgell wrote:On July 07 2019 08:41 travis wrote: The easiest way to put it is that the constructed polgyon is the one with the maximum size, and that the original lines can be included within the bounds of the polygon (for example if the line criss crossed such that they made a shape like a plus sign or a cross, then the polygon would be the diamond around that cross) That is not guaranteeing a unique solution. Can you give me an example of why not?
Offhand, imagine the following scenario I whipped up in paint:
The two line segments are perpendicular, and C and D are equidistant from B. Therefore A is equidistant from C and D as well. Do you connect AC and BD, or AD and BC?

United States23581 Posts
Ah, maybe I miscommunicated that all endpoints needed to be along the boundary of the polygon. They don't. In that example, we would construct the triangle AC,CD,DA

Okay, so we are simply talking about the convex hull of the 4 input points?
And let's just never mention those two lines again, because they really also don't seem to matter here.
Okay.... so now that it seems like we finally figured out the task.... for your other problem in the most generic way and compatible with the x/y description you gave originally:
As your shape is convex, you can simply describe the inside of your polygon by 3 or 4 inequalities (one for each side), so you only have to check if your point of question is fulfilling those.
So if we are talking about the most brute forcish algorithm possible: (the final states are bolded) 1) You loop over all possible unordered pairs of points from your 4 points. 1a) There are no pairs left: You are done. The point is inside the polygon 1b) There are still pairs left: > 2) 2) You calculate the line through those points 3) You check on which side of your line the other two original points are. 3a) They are on different sides > go back to 1) 3b) They are on the same side > 4) 4) You check if your point of question is on the same side: 4a) yes > go back to 1) 4b) no > abort, the point is not inside your polygon
The cases where a point is on the line in 3) are trivial, in the case of the point being on the line in 4) it really depends on how you define "in the polygon"

United States23581 Posts
number theory question:
we know 12^10 = 61917ab4224 (where a and b are digits)
find a and b using congruences mod 9, and mod 11.
how the f do I do that? I know ab is a multiple of 12..

I would start with calculating 12^10 mod 9 and mod 11, possibly using the methods we talked about a while back in this thread.
+ Show Spoiler +It turns out that these results are pretty easy to calculate, as 12 = 3 mods 9, and 3^2 = 0 mod 9, so 12 ^even number = 0 mod 9. Similarly, 12 = 1 (mod 11), so 12 ^even number = 1 (mod 11)
Since 61917ab4224 = 12^10, this equality must also be true for the mod results. Then try to figure out what you can say about 61917ab4224 (mod 9) or (mod 11), and see if this leads you to a conclusion.
+ Show Spoiler +So 61917ab4224 = 0 (mod 9) and 61917ab4224 = 1 (mod 11). You can use the sum of the digits to deal with the (mod 9) stuff. 6+1+9+1+7+4+2+2+4+ a + b must equal 0(mod9), because if 61917ab4224 is dividable by 9, which it is, then it's digit sum must be a multiple of 9, so 0(mod9). And since 6+1+9+1+7+4+2+2+4 =36 = 0, this means that a+b = 0 (mod 9). If you can actually prove that ab is a multiple of 12, this means that ab must be 36 or 72. Otherwise, it could also be 18, 27, 36, 45, 54, 63, 72, 81, 90, 99. But we have greatly reduced the amount of possible combinations here.
So lets look at the (mod 11) stuff now. 61917ab4224 = 1 (mod 11). I actually have no idea what to expect here or how to deal with this, so i will just write what i do, and we will see if something useful is the result. We can remove any multiple of 11 from the right side of this equation. So 61917ab4224 = 1 (mod 11) = 50917ab4224 = 06917ab4224 = 00317ab4224 = 00097ab4224 = 00009ab4224 = 9ab4224 = 9ab2002=9ab0902 = 9ab0022 = 9ab0000
So 9ab0000=1 (mod11). This means directly that 9ab can not be dividable by 11 (not that interesting, but it kicks out a few of our possibilities. since 990 = 0 (mod11), this means ab cannot be 90. 79, 68,57,46,35,24,13 are all not on our list of possible results, so we don't really gain anything interesting.
I get the feeling that maybe 9ab must also be equal to 1 (mod 11), that would directly lead us to 36 being the result (which a quick check of 12^10 with the calculator shows me is true), but i am not quite certain why that would be the case. Can we just get rid of zeros at the end of the number (clearly not true in general for mod calculation, as 50(mod 6) = 2, while 5 (mod6) = 5, but maybe it is true for (mod11), or for (mod prime)? might be that there is a theorem for this, but i don't remember). (Also not true for mod prime, as 50 (mod 7) = 1, while 5 mod 7 = 2
So lets think about this. If a<b, we can remove aa0000 as a multiple of 11 from 9ab0000. This means that 90(ba)0000 = 1, but this doesn't seem fruitful.
AHA! 10 = 1 (mod 11), so dividing by 10 is the same as dividing by 1 (I am pretty sure that i can divide in mod 11 like this). This means that 9ab0000/10000 = 1/(1)^4 ===> 9ab = 1(mod 11), which is true for 991, 980, 969, 958, 947, 936, 925, 914, 903. And that means that ab is element of {91, 80, 68,58,47,36, 25, 14, 03}. But we already know from the first part that ab is elemnt of {18, 27, 36, 45, 54, 63, 72, 81, 90, 99}. The only overlap between the two is 36, so ab must be 36.
(Turns out the above is actually incorrect, i looked it up and i cannot generally divide like that in modulo operations. However, i can divide if the greatest common denomitor of the number i want to divide by (10) and the modulo number (11) is 1, which it is)
I hope that this helps with how one can approach such a problem without actually knowing what to do, or remembering a lot of the relevant theorems. There is probably a way simpler solution for the mod 11 stuff, but since i have no idea what that is, i choose to rather do an ugly sloppy solution than none.

What are some real life examples where data makes a perfect parabola?

What do you mean with "perfect parabola"? In realworld situations, you never get a "perfect" something. There is always some variance in your data.
Also, what kind of "realworld" situations are you looking for? Is free fall outside of an atmosphere okay?

On August 13 2019 01:23 Starlightsun wrote: What are some real life examples where data makes a perfect parabola?
 The braking distance of cars follows a quadratic equation, which is useful in the contexts of traffic reports, the DMV, and car accident investigations.  Professional landscaping and optimizing the positions of sprinkler systems in gardens and lawns for efficient watering use quadratics, since the water’s trajectory is always an arc (a parabola).  Civil engineers and architects frequently design and build structures that require parabolas, such as tunnels, bridges, dams, and buildings.  Any sort of projectile or projectile motion (trajectory) involves parabolas, such as: throwing darts, firing arrows, shooting bullets, throwing or kicking or hitting balls, launching rockets, firing cannonballs, juggling objects, jumping off ski jumps or bike jumps, horseback riding and jumping, dolphins jumping, riding snowboards and skateboards off halfpipes, etc.  Parabolic surfaces are used to magnify acoustics in auditoriums and speakers.  Parabolic surfaces are used in optics to optimize vision and visual displays, such as with camera lenses, lenses of glasses, and contact lenses.  Parabolic surfaces are used to generate, focus, and transfer heat, light, and energy in heaters, headlights, satellite dishes, and other devices.  Many roller coaster tracks and other amusement park rides are parabolic for safety and speed reasons.

Thanks DarkPlasmaBall. That's fascinating how many applications there are. I'm learning quadratics in algebra and just wondered what they're used to model.



