# GRR: Need more math help!

Discussion in 'Community' started by springscansing, Jun 6, 2003.

1. ### springscansing macrumors 6502a

Joined:
Oct 13, 2002
Location:
New York
#1
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.

Joined:
Oct 13, 2002
Location:
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3. ### Mr. Anderson Moderator emeritus

Joined:
Nov 1, 2001
Location:
VA
#3
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

4. ### szark macrumors 68030

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Arid-Zone-A
#4
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.

5. ### Veldek macrumors 68000

Joined:
Mar 29, 2003
Location:
Germany
#5
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.

6. ### Mr. Anderson Moderator emeritus

Joined:
Nov 1, 2001
Location:
VA
#6
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

7. ### Veldek macrumors 68000

Joined:
Mar 29, 2003
Location:
Germany
#7
#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.

8. ### szark macrumors 68030

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May 14, 2002
Location:
Arid-Zone-A
#8
From a Math Forum:

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

9. ### Mr. Anderson Moderator emeritus

Joined:
Nov 1, 2001
Location:
VA
#9
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

### Staff Member

Joined:
Sep 19, 2002
Location:
Los Angeles
#10
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.

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

11. ### springscansing thread starter macrumors 6502a

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Oct 13, 2002
Location:
New York
#11
I thought 1 was a prime number.. doh.

That's what was throwing me off.

Thanks for the help!

12. ### NicoMan macrumors 6502a

Joined:
Oct 20, 2002
Location:
Malmö, Sweden
#12
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

### Staff Member

Joined:
Sep 19, 2002
Location:
Los Angeles
#13
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.

14. ### sparkleytone macrumors 68020

Joined:
Oct 28, 2001
Location:
Greensboro, NC
#14
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.

15. ### bennetsaysargh macrumors 68020

Joined:
Jan 20, 2003
Location:
New York
#15
what is the correct answer to #14? i just wanna see if i ogt it right

16. ### sparkleytone macrumors 68020

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Oct 28, 2001
Location:
Greensboro, NC
#16
i slightly edited it for clarity, but that looks right to me.

17. ### bennetsaysargh macrumors 68020

Joined:
Jan 20, 2003
Location:
New York
#17
yaty! i got it right! should it have been that easy? (#14)

18. ### sparkleytone macrumors 68020

Joined:
Oct 28, 2001
Location:
Greensboro, NC
#18
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.

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Jan 20, 2003
Location:
New York
20. ### beez7777 macrumors 6502a

Joined:
Aug 5, 2002
Location:
Notre Dame
#20
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.

21. ### bennetsaysargh macrumors 68020

Joined:
Jan 20, 2003
Location:
New York
#21
so would i be able to take it this fall as a freshman? or is there a certain age that you have to be?>

22. ### sparkleytone macrumors 68020

Joined:
Oct 28, 2001
Location:
Greensboro, NC
#22
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.

23. ### janey macrumors 603

Joined:
Dec 20, 2002
Location:
sunny los angeles
#23
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 and the Fibonacci sequence: If you add up the numbers in Pascal's triangle diagonally, you get the fibonacci numbers.
It's really cool the fibonacci numbers are everywhere!

### Staff Member

Joined:
Sep 19, 2002
Location:
Los Angeles
#24
Well, not ALL of them! The ellipsis after 34 was literal. But it was cool, as you say.
And it's fun to prove those facts by induction (at least if you are a math fancier like me).
Actually, it's one less, not one more.

25. ### janey macrumors 603

Joined:
Dec 20, 2002
Location:
sunny los angeles
#25
oops