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SAT math question

Started by Sr_Guapo Oct 9, 2004 at 10:08 PM 29 replies 3.9k views

Original Post

Oct 9, 2004 at 10:08 PM

Ok, I just had the SAT I this morning and I have a question about one of the problems. Here it is: Say n=a^2*b*c, where a, b and c are different prime numbers greater than two. How many integers greter than 2 and less than n are divisors of n? WTF? I had no idea how to do it mathmetically, so I started plugging and chugging (it was the last question and I had 10 minutes left). I used 3, 5, and 7 (respectfully), and got n=315. I then found all the divisors of 315: 3,5,7,9,15,21,35,45,63,105: so that is ten numbers (I actually got 8 on the test, whatever). Anyway, one friend of mine got 7, one got 9, another 10, and everyone else got something else. I tried it again with a different set of numbers, and I got a completey different answer. Did I miss something in the problem or what? Also, more importantly, how do you solved this mathematically (not gues and check)? I posted this in the lounge because I wanted to get a fast response, but feel free to move it to the Math/ Physics forum if neccessary. Thanks alot!

Tyro

Oct 9, 2004 at 10:14 PM

permutations and combinations. I forgot how exactly you can calculate using that (ill look into my old math notes)

capn_midnight

404

capn_midnight

Oct 9, 2004 at 10:16 PM

Quote:
Original post by Sr_Guapo
Ok, I just had the SAT I this morning and I have a question about one of the problems. Here it is:

Say n=a^2*b*c, where a, b and c are different prime numbers greater than two. How many integers greter than 2 and less than n are divisors of n?

WTF? I had no idea how to do it mathmetically, so I started plugging and chugging (it was the last question and I had 10 minutes left). I used 3, 5, and 7 (respectfully), and got n=315. I then found all the divisors of 315:

3,5,7,9,15,21,35,45,63,105: so that is ten numbers (I actually got 8 on the test, whatever). Anyway, one friend of mine got 7, one got 9, another 10, and everyone else got something else. I tried it again with a different set of numbers, and I got a completey different answer.

Did I miss something in the problem or what? Also, more importantly, how do you solved this mathematically (not gues and check)?

I posted this in the lounge because I wanted to get a fast response, but feel free to move it to the Math/
Physics forum if neccessary. Thanks alot!

n=a^2*b*c. a, b, c are prime.
For any a, b, or c, there are only 7 integers that are divisors greater than two and less than n.

a, b, c, a*b, a*c, b*c, a*b*c

edit: I just reallized, it actually took me longer to type the responce than to figure out the problem.

Sr_Guapo

Author

Sr_Guapo

Oct 9, 2004 at 10:18 PM

I understand Permutations and Combinations, but I don't see how those fit in. I tried doing it like:

A prime number times itself gives a number that has 1 factor. A prime number times another prime number give a number with two factors. I then couldn't figure out what would happen if you multiply a number with 2 factors with a number with 1 factor (I was running out of time).

Sr_Guapo

Author

Sr_Guapo

Oct 9, 2004 at 10:20 PM

Quote:
Original post by capn_midnight

n=a^2*b*c. a, b, c are prime.
For any a, b, or c, there are only 7 integers that are divisors greater than two and less than n.

Can you explain how you came to that conclusion? I am still not understanding it...

capn_midnight

404

capn_midnight

Oct 9, 2004 at 10:24 PM

Quote:
Original post by Sr_Guapo
Quote:
Original post by capn_midnight

n=a^2*b*c. a, b, c are prime.
For any a, b, or c, there are only 7 integers that are divisors greater than two and less than n.

Can you explain how you came to that conclusion? I am still not understanding it...

the numbers are prime, so they can't be factored. Since they are prime, then they are integers, and are therefore obviously factors of n. You are looking for unique permutations of a, a, b, and c that are greater than 2 and less than n. Since a, b, and c are all greater than 2, we don't have to worry about the former restraint. Since a*a*b*c=n, then that is the only permutation we need to elliminate. That leaves
a
b
c
a*a
a*b
a*c
b*c
a*a*b
a*a*c
a*b*c

ten, actually. I don't know why I said 7.

Sr_Guapo

Author

Sr_Guapo

Oct 9, 2004 at 10:27 PM

Quote:
Original post by capn_midnight
Quote:
Original post by Sr_Guapo
Quote:
Original post by capn_midnight

n=a^2*b*c. a, b, c are prime.
For any a, b, or c, there are only 7 integers that are divisors greater than two and less than n.

Can you explain how you came to that conclusion? I am still not understanding it...

the numbers are prime, so they can't be factored. Since they are prime, then they are integers, and are therefore obviously factors of n. You are looking for unique permutations of a, a, b, and c that are greater than 2 and less than n. Since a, b, and c are all greater than 2, we don't have to worry about the former restraint. Since a*a*b*c=n, then that is the only permutation we need to elliminate. That leaves
a
b
c
a*a
a*b
a*c
b*c
a*a*b
a*a*c
a*b*c

ten, actually. I don't know why I said 7.

Thanks alot, I think I understand now... Damn it, though, I narrowed it down to 8 (must have missed a couple) and 10, and I guessed 8 at the last second... Oh well...

BTW, one of my friends explained it like you did at first, that is why he got 7.

choffstein

117

choffstein

Oct 9, 2004 at 10:28 PM

It's cool. I thought it was a trick question and guessed 3 =D

Tyro

Oct 9, 2004 at 10:32 PM

i can come to the same conclusion mathematically, just wait till i whip out my notes :P

trial and error is for wussies:P
im joking, thats how i solved it too anyways, but if there was, instead of 3 prime numbers, 123091283 prime numbers as factors, the mathematical way is more useful.

my notes are stashed somewhere in the basement or something maybe i can give you an answer by tomorow.

Iam sure permutation is the way to solve this because i remember solving problems like this in my math class.

choffstein

117

choffstein

Oct 9, 2004 at 10:34 PM

what math was this taught in?

I find the math to be hard despite the fact im learning Calculus now because I learned it so long ago. I barely remember my geometry. The only reason I know the stuff is because I figure it out during the test itself.

capn_midnight

404

capn_midnight

Oct 9, 2004 at 10:59 PM

Quote:
Original post by visage
what math was this taught in?

I find the math to be hard despite the fact im learning Calculus now because I learned it so long ago. I barely remember my geometry. The only reason I know the stuff is because I figure it out during the test itself.

combinatronics, or discrete mathematics.

Promit

Mod

1,082

Promit

Oct 9, 2004 at 11:01 PM

FYI, it's not actually legal to post this [lol]

Sr_Guapo

Author

Sr_Guapo

Oct 10, 2004 at 12:02 AM

Quote:
Original post by Promit
FYI, it's not actually legal to post this [lol]

Really? Why is that? Is it copyrighted material or something? If a mod agrees, feel free to close the post, I had no idea...

Promit

Mod

1,082

Promit

Oct 10, 2004 at 12:16 AM

Meh, nobody pays attention to their rules, but technically speaking, you're not allowed to discuss the multiple choice with anyone ever or you die.

JoHnOnIzEr

Oct 10, 2004 at 12:47 AM

yay i got at least one right on the SATs today! but i already got an 800 before :(

Zahlman

257

Zahlman

Oct 10, 2004 at 2:13 AM

A divisor of n, including 1 and n, is of the form a^x b^y c^z, where x in 0..2, y in 0..1, z in 0..1.

That makes 3 * 2 * 2 = 12 combinations.

We disallow 1 and n, yielding 10. (We do not need to disallow 2, because a,b,c are given to be > 2, implying n is odd.)

Code_Dark

103

Code_Dark

Oct 10, 2004 at 2:45 AM

Quote:
Original post by JoHnOnIzEr
yay i got at least one right on the SATs today! but i already got an 800 before :(

Out of 1600? Why, that's only 50%! [grin]

etothex

Oct 10, 2004 at 2:56 AM

Quote:
Original post by Sr_Guapo
Quote:
Original post by Promit
FYI, it's not actually legal to post this [lol]

Really? Why is that? Is it copyrighted material or something? If a mod agrees, feel free to close the post, I had no idea...

Quote:
Original post by Promit
Meh, nobody pays attention to their rules, but technically speaking, you're not allowed to discuss the multiple choice with anyone ever or you die.

Well, that's for the AP exams, where you sign a contract not to reveal the multiple choice questions ever. That's because they recycle the multiple choice on the AP exams, because they need to ensure absolute fairness (and the chances of one person taking the same AP exam twice are minimal, while people routinely take the SAT I twice)

Therefore there's a much larger question pool for the SAT I, so it's not so much a question of cheating (which sharing AP mc questions automatically is) but rather, like you said, posting a copyrighted question. While it isn't as serious as cheating, they try to discourage it.

Ilici

205

Ilici

Oct 10, 2004 at 3:22 AM

I took the SATs too yesterday, and i think i answered 9 at that question, going on the same logic that capn_midnight explained but i think i forgot one factor. fuck...

Dmytry

264

Dmytry

Oct 10, 2004 at 9:05 AM

eh, too late to solve. Anyway, simple, as you know the factors, and then just do as Zahlman said, no need to think at all, no way to make mistakes.
Anyway, WTF is SAT?

Quote:
Original post by etothex
Quote:
Original post by Sr_Guapo
Quote:
Original post by Promit
FYI, it's not actually legal to post this [lol]

Really? Why is that? Is it copyrighted material or something? If a mod agrees, feel free to close the post, I had no idea...

Quote:
Original post by Promit
Meh, nobody pays attention to their rules, but technically speaking, you're not allowed to discuss the multiple choice with anyone ever or you die.

Well, that's for the AP exams, where you sign a contract not to reveal the multiple choice questions ever. That's because they recycle the multiple choice on the AP exams, because they need to ensure absolute fairness (and the chances of one person taking the same AP exam twice are minimal, while people routinely take the SAT I twice)

Therefore there's a much larger question pool for the SAT I, so it's not so much a question of cheating (which sharing AP mc questions automatically is) but rather, like you said, posting a copyrighted question. While it isn't as serious as cheating, they try to discourage it.

And what, they use same questions on some exams over and over? What's fair with it? Any information leaks like hell no matter what people sign, no matter if they swear on Bible or not. Also, can a <18 childs sign a contract?
What is "AP exams"?

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