"How do I figure out the math q’s?"

[master_q 0]

"How do I figure out the math q’s?"

"Master Q gave in the chat room . . ?"

 

 

Some of you were saying that you did not even know how to approach the problem and many of you had troubles trying to figure them out. Maybe on one of the questions I probably should have waited a minute or so, but oh well.

 

The first question was:

Q7: In the science board I did a topic on prime #s so I’ll ask a prime # question. What is the first pseudo-prime #? What is a pseudo-prime?

 

To get some info on primes and pseudo-primes you can read the topic that’s dedicated for prime #s in the science board. Well, to answer in brief basically a pseudoprime is a # that is not a prime, but obeys Fermat’s property.

 

Fermat’s property says that:

If any # p is a prime (or pseudoprime) and any # a < p, then a^(p-1) - 1 will be divisible by p. This means that if we have a # and we want to see if it passes this test then we would place a value less then that # for a and put the # in for p, then you figure it out. After that you see if you can divide it by p. If it is divisible (meaning if it comes out whole with a remainder of zero), then we would have passed the test. Like I said this is to test primes, but every in a while you find a composite works with this test. These composites are called pseudoprimes. The first pseudoprime is 341.

 

Q11: A spring’s first loop is 20 inches long. Its second loop is 9/10 of this length; the third is 9/10 of the second; ect. Suppose the spring has infinitely many loops. What is its length (if possible)?

 

Well this one is not as hard as it looks. The concept or proof might be a bit more complex to understand, but if you know the general equations you can figure it out. I’m surprised that no one said; “it’s not possible,” because many would assume that infinitely means it would go on forever. However, this is not the case and I can give a easy example.

 

Draw a big square on a piece of paper.

Draw a line to cut that in half. Color one of the half’s.

Now look at the half that is not colored. Draw a line and cut that half in half. Color one of those half’s.

Continue to do this . . . . . . .

 

This shows that mathematically speaking we could cut it in half forever (but of course physics puts a limit on that. However, that’s a different story). We can always divide something by 2. We can continue to do this . . . . . .we can do this forever. However, if you did this for a while numerically and watched, then you would see it approach something. If anyone has read any of my editions to Star Trek & Physics Weekly and have read the editions that were on logarithmic equations, then you know when we graph them on the x,y plane that they approach a asymptote. The function will never get there, but it will get closer and closer. It is basically (to a degree) the same here and that sort of logic and be applied here. So if we are dividing something like we are now, then we can compute that if we have a series like we have now (it’s geometric) and if our constant that we keep on multiplying is the same*, then we can compute the summation or sum of all of the series.

 

*Not to get anyone confused, but it might be a bit confusing when I say multiply or divide as I am saying. However, we are multiplying by (1/2) which is the same as dividing by 2. In standard form when writing these kinds of problems out you traditionally will write them out as multiplication by a fraction which would be equivalent to dividing by the equivalent # or variables at hand.

 

Now in this case we can follow a simple equation: (This however, demotes our lovely sigma or greek letter in summation) . . . . . . (a₁) / (1 - r) = sum

That’s it. 20/(1 – (9/10)) = 200 inches

 

Q15: A Starfleet Shuttle Craft is moving at a constant rate of 20m/s and passes a starship that is not moving. The starship waits 3 seconds, then pursues the shuttle craft, accelerating at a constant rate of 4 m/s^2.

 

Well there are lots and lots of ways we could solve this. I will just look at one that probably is the easiest . . . You have to figure out where they intercept. The key thing is that both will intercept when they have the same displacement. For a result you can set the shuttle’s displacement and the starship’s displacement equal to one each other.

 

Let’s first write two general equations that describe the motion of the two:

delta x = v(t+3) shuttle

delta x = 1/2at^2 starship

 

v(t+3) = 0.5at^2

(20 m/s) (t + 3) = 0.5 (4 m/s^2)t^2

20t + 60 = 2t^2

2t^2 - 20 t - 60 = 0

 

Remember? The quadratic formula?

x = { -b plus or minus (b^2 - 4 ac)^(1/2) } /(2a)

 

t = { 20 plus or minus (400 + 480)^(1/2) } /(4)

. . .

t = 12.4s

So after the starship has accelerated 12.4s, he catches the speeder who has been traveling for 15.4s.

 

Q19: A “perfect #” is a # that all of its factors add up to that #

(Factor: It’s simply a # that can be divided evenly into another number)

For example 6 is a perfect # because all of its factors add up to it! (1+2+3=6)

Find the second perfect #!

Now a question like this I will probably ask again. In our next session I won’t really have that much math and I won’t have any physics, but a question like this I think is simple and nice. Anyone could figure a problem like this one out very easily. Math will be still there, but I’ll probably do something like astronomy next session instead of plan physics.

 

Anyways, to approach this question you have to just figure out the factors of some #s and add them up . . . (well basically).

 

I gave a hint: I said it was between some #s. Let’s say that the hint was/is: 26 to 28

 

Factors of 28: 1, 2, 4, 7, 14

(remember 1,2,4,7,14 are factors because they divide evenly as in you get a whole #)

1 + 2 + 4 + 7 + 14 = 28

It is a perfect #!

 

 

It’s not hard.

 

 

For our next session (hopefully some more will join) I’ll probably only post one or two math questions at the most.

 

 

Master Q

StarTrek_Master_Q@yahoo.com