View Full Version : GRR: Need more math help!

springscansing

Jun 6, 2003, 09:49 AM

Sorry about this again, heh. Could use some help on 2 questions for my SAT review here:

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#14:

When the square of 2m is multiplied by 2, the result is g.

m > 0

For the two quantities below is A greater, B greater, both the same, or is the relationship indeterminable:

g/4m -- m

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#15

p and r are different prime numbers.

For the two quantities below is A greater, B greater, both the same, or is the relationship indeterminable:

The number of positive integer divisors of p^3 -- The number of positive integer divisors of pr

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Now I got answers for both of them, but they're not what the book says is right, but I am almost positive I'm correct. For #14, I got 'm' as greater, and for #15, I got 'indeterminable'. The book says for #14 that g/4m is greater, and for #15 that they're equal.

If someone can either tell me the book is full of crap or why I'm wrong I'd GREATLY appreciate it.

springscansing

Jun 6, 2003, 10:00 AM

Nevermind #14, just look at #15.

Mr. Anderson

Jun 6, 2003, 10:46 AM

using 1 and 2 as the two prime numbers gets different postive integer divisors than using 5 and 7.

That to me proves that its indeterminable - unless I too am missing something.

D

szark

Jun 6, 2003, 11:44 AM

Just had to point out to D that 1 isn't a prime number... :)

Having said that, I have no idea what the answer is.

Veldek

Jun 6, 2003, 12:07 PM

Hello, I studied math and I think I got the solution to #14:

2*(2*m)^2 = g

<=> 8*m^2 = g

<=> g/(4*m) = 2*m > m

because m > 0.

I will now look at #15.

Mr. Anderson

Jun 6, 2003, 12:18 PM

Originally posted by szark

Just had to point out to D that 1 isn't a prime number... :)

Having said that, I have no idea what the answer is.

ah, really - its been a while, but I thought the definition of a prime was its only divisible by 1 and itself - the number 1 fits that definition.

In that case, take 2 and 3

2^3= 8 --> 1,2,4,8

3^3= 27 -->1,3,9,27

5^3=125 -->1,5,25,125

7^3= 343 --> 1, 7, 49, 343

2*3 = 6 --> 1,2,3,6

2*5 = 10 --> 1,2,5,10

3*5 = 15 --> 1,3,5,15

etc.

They're equal - even though you're cubing the prime, its only got 4 divisors (on of them being the square of the prime and itself)

Multiplying to primes you only get 4 as well, 1, the result and the two primes.

D

Veldek

Jun 6, 2003, 12:20 PM

#15:

p^3 has the following positive integer divisors:

1, p, p^2, p^3

as the only positive integer divisors of p are 1 and p.

p*r has the following positive integer divisors:

1, p, r, p*r

as the only positive integer divisors of p are 1 and p and of r are 1 and r.

Well, I'm not sure if one can call this a proof, but it will perhaps help you a little.

szark

Jun 6, 2003, 12:36 PM

Originally posted by Mr. Anderson

ah, really - its been a while, but I thought the definition of a prime was its only divisible by 1 and itself - the number 1 fits that definition.

From a Math Forum:

A prime number is a positive integer that has exactly two positive integer factors, 1 and itself.

Note that the definition of a prime number doesn't allow 1 to be a prime number: 1 only has one factor, namely 1. Prime numbers have exactly two factors, not "at most two" or anything like that.

I'm not a math geek or anything, just trying to help. :D

Mr. Anderson

Jun 6, 2003, 12:40 PM

Originally posted by szark

I'm not a math geek or anything, just trying to help. :D

Oh that's fine, I'm not debating - as you can see on my brute force approach to the question above ;)

I got the same results, but much less elegantly than Veldek. It was the 1 that was making me think things were other than equal.

D

Doctor Q

Jun 6, 2003, 12:45 PM

The positive integer divisors of p^3:

1

p

pp

ppp

To count them, you have to remove the duplicates. But since p is prime, p > 2, so there are no cases where any of these values are the same.

Therefore, A (the number of positive integer divisors of p^3) = 4.

The positive integer divisors of pr:

1

p

r

pr

To count them, you have to remove the duplicates. But since p and r are prime and you are told they differ, there are no cases where any of these values are the same.

Therefore, B (the number of positive integer divisors of pr) = 4.

Answer: A=B

Note that the answer would be different if they hadn't told you that p and r are different prime numbers!

springscansing

Jun 6, 2003, 01:49 PM

I thought 1 was a prime number.. doh.

That's what was throwing me off.

Thanks for the help!

NicoMan

Jun 6, 2003, 02:16 PM

You guys are missing something: all those proofs you have stated are based on one important fact that it might be important to state in your solution (I don't know how it works in your tests, but I know that when I studied you had to explain your answers). The decomposition of an integer in prime numbers is UNIQUE.

Hope you get my drift.

NicoMan

Doctor Q

Jun 7, 2003, 01:59 AM

There must be math vibes in the air today. First there was this thread. Then this evening, when I was driving, the license plate frame of the guy in front of me said FIBONACCI at the top and 1 1 2 3 5 8 13 21 34... at the bottom. I wanted to ask him why he had the Fibonacci series on his license plate frame but I couldn't get next to him at a red light. Now we'll never know.

sparkleytone

Jun 7, 2003, 08:42 PM

Originally posted by Veldek

#15:

p^3 has the following positive integer divisors:

1, p, p^2, p^3

as the only positive integer divisors of p are 1 and p.

p*r has the following positive integer divisors:

1, p, r, p*r

as the only positive integer divisors of p are 1 and p and of r are 1 and r.

Well, I'm not sure if one can call this a proof, but it will perhaps help you a little.

this guy is correct. you can view the prime number multiplication as a tree...kinda like this.

5 -> p

25 -> p^2

125 -> p^3

625 -> p^4

3125 -> p^5

etc. where p > 0 and p is a subset of the integers and p is only divisible by p and 1. sorry my math symbols fail me, its been a WHILE. the resulting p^x are only divisible by itself (p^x), every number below (above) it on the tree, and 1. this is inherited by the prime funtion of 'p'. really in order to do a good solid proof, you'd need to use logarithmic functions which IIRC arent covered for the SAT.

bennetsaysargh

Jun 7, 2003, 09:52 PM

what is the correct answer to #14? i just wanna see if i ogt it right:D

sparkleytone

Jun 7, 2003, 11:11 PM

Originally posted by Veldek

Hello, I studied math and I think I got the solution to #14:

2(2m)^2 = g

<=> 8m^2 = g

<=> g/4m = 2m

2m > m

because m > 0.

i slightly edited it for clarity, but that looks right to me.

bennetsaysargh

Jun 8, 2003, 09:33 AM

yaty! i got it right! should it have been that easy? (#14)

sparkleytone

Jun 8, 2003, 10:30 AM

Originally posted by bennetsaysargh

yaty! i got it right! should it have been that easy? (#14)

lol, yeah. its the SAT man, it ain't hard. i messed up by going in and thinking too much. just remember its a test, and a standardized test. learn the test and you'll get a higher score. learn the material and you'll actually learn.

bennetsaysargh

Jun 8, 2003, 11:08 AM

im only in 8th grade, what grade is SAT's?

beez7777

Jun 8, 2003, 11:29 AM

Originally posted by bennetsaysargh

im only in 8th grade, what grade is SAT's?

you can take the SAT's whenever you want. i'm in 10th grade and i took them this spring, but i'll probably take them a few more times over the next 2 years.

most people take the SAT's in 11th grade, and then if they want to, again whenever they want.

bennetsaysargh

Jun 8, 2003, 11:36 AM

so would i be able to take it this fall as a freshman? or is there a certain age that you have to be?>

sparkleytone

Jun 8, 2003, 12:39 PM

there is really no point in taking it until your junior yr. its only something for colleges to look at. taking it earlier is basically just a way to get bragging rights.

janey

Jun 8, 2003, 02:07 PM

re: one as a prime number

in algebra my math teacher would tell me that "0" and "1" are nerds, and that they're not prime. She also said for those geeks out there, the two numbers you use in binary (base 2), are not prime.

doctor q: did that guy really have the fibonacci sequence on this licence plate frame? that's so awesome

here are some interesting facts about the fibonacci sequence:

1. if the nth number in the sequence is a, then every nth number after that is a multiple of a

2. take any number in the sequence and square it, the result will always be one more than the product of the previous number and the following number in the sequence

3. take any number in the sequence and the third number after that, add then and the sum will be exactly twice the second number between them in the sequence.

4. take any number in the sequence and the fourth number after that, add them and the sum will always be 3 times the number halfway between them in the sequence.

5. add up all the numbers in the sequence up to any point, and the sum will be one less than the second number yet to come.

The Connection between Pascal's Triangle (http://ptri1.tripod.com/) and the Fibonacci sequence: If you add up the numbers in Pascal's triangle diagonally, you get the fibonacci numbers.

It's really cool :p the fibonacci numbers are everywhere!

Doctor Q

Jun 8, 2003, 04:32 PM

Originally posted by übergeek

doctor q: did that guy really have the fibonacci sequence on this licence plate frame?Well, not ALL of them! :p The ellipsis after 34 was literal. But it was cool, as you say.here are some interesting facts about the fibonacci sequence...And it's fun to prove those facts by induction (at least if you are a math fancier like me).2. take any number in the sequence and square it, the result will always be one more than the product of the previous number and the following number in the sequenceActually, it's one less, not one more.

janey

Jun 8, 2003, 04:56 PM

Originally posted by Doctor Q

...Actually, it's one less, not one more.

http://community.the-underdogs.org/smiley/misc/ashamed.gif oops :p

Doctor Q

Jun 8, 2003, 05:56 PM

For more amazing math, check out Surreal numbers (http://www.wikipedia.org/wiki/Surreal_number), invented by John Conway. I learned about them in a book by Martin Gardner.

A surreal number is an ordered pair of sets such that all members of the 2nd set are greater than all members of the 1st set. If you call the pair of sets [ {} , {} ] by the name "0", the pair of sets [ 0, {} ] by the name "1", the pair of sets [ 1, {} ] by the name "2", etc., then with appropriate definitions of addition and multiplication you find that all the usual properties of arithmetic apply. Same for subtraction, division, reciprocals, etc. With other combinations of left and right sets, you can define the power-of-two rationals, then all rationals, reals, infinitesimals, infinite sets, and lots of other categories. And all this starting from "nothing" (a pair of empty sets) and applying a few simple rules!

If any of you like the beauty of math, take a peek. If you believe, like Mattel's Teen Talk Barbie (http://www.sniggle.net/barbie.php), that math is no fun, or you are busy studying algebra for the S.A.T., save it for a rainy day.