Math question.

2006-10-18 16:31:57

#1

Goven: /̵͇̿̿/'̿'̿ ̿; +125|6477|Purdue

I took the PSATs today and there was a problem (actually a couple) in the math section that I couldn't figure out for the life of me. It said
"k, x, and a are positive integers. What does k-a equal if
kxa=x^a+1"

2006-10-18 16:33:00

#2

Ryan: Member; +1,230|6840|Alberta, Canada

...Little late to be asking

I'm only in grade 9 so I wouldn't know, and I don't have math this semester.

Last edited by ryan_14 (2006-10-18 16:33:22)

2006-10-18 16:35:40

#3

Goven: /̵͇̿̿/'̿'̿ ̿; +125|6477|Purdue

I'm only a sophomore :p. Only reason I'm asking is because it's been bugging me all day.

2006-10-18 16:37:45

#4

-TL-: Srs lurker; +25|6489|Oklahoma City

is it x^a+1(k-a)/kxa ?

Last edited by -TL- (2006-10-18 16:41:08)

2006-10-18 16:38:57

#5

Varegg: Support fanatic :-); +2,206|6807|Nårvei

http://www.mathleague.com/help/integers/integers.htm

Mathleague has the answer !

Wait behind the line ..............................................................

2006-10-18 16:39:26

#6

Goven: /̵͇̿̿/'̿'̿ ̿; +125|6477|Purdue

-TL- wrote:
is it x^a+1(k-a)/kxa ?

No, the question wants the answer for k-a.
You have to find out what k, a, and x equal with kxa=x^a+1

Varegg wrote:
http://www.mathleague.com/help/integers/integers.htm

Mathleague has the answer !

And...no...it doesn't.

Last edited by Goven (2006-10-18 16:44:16)

2006-10-18 16:42:11

#7

evilcartman99: The Octagon; +18|6409|da ville, va

You actually took the PSAT's, why? I'm a sophmore too but I don't plan on taking the PSAT's but I might consider it next year. Would it be Kx^a+1)/(xa)

Last edited by evilcartman99 (2006-10-18 16:43:56)

2006-10-18 16:43:01

#8

=Robin-Hood=: A stranger in the dark; +213|6817|Belgium

In a quick glance

k-a = (x^a-a²)/a

But my guess is you should be able to simplify that

To late and to long ago.

Cheers,
R

2006-10-18 16:45:00

#9

Goven: /̵͇̿̿/'̿'̿ ̿; +125|6477|Purdue

=Robin-Hood= wrote:
In a quick glance

k-a = (x^a-a²)/a

But my guess is you should be able to simplify that

To late and to long ago.

Cheers,
R

Unfortunately, that wouldn't work. It was a gridded response problem, and it actually want integers as an answer.

evilcartman99 wrote:
You actually took the PSAT's, why? I'm a sophmore too but I don't plan on taking the PSAT's but I might consider it next year. Would it be Kx^a+1)/(xa)

It's required for sophomores where I go to school. It's optional for Juniors though.

Last edited by Goven (2006-10-18 16:46:01)

2006-10-18 16:46:37

#10

-TL-: Srs lurker; +25|6489|Oklahoma City

-TL- wrote:
is it x^a+1(k-a)/kxa ?

What I meant, was
if kax=x^a-1
then k = x^a-1/ax
and a = x^a-1/kx
so k - a = x^a-1/ax - x^a-1/kx = x^a+1(k-a)/kxa
But if the test wanted an integer as an answer, I have no idea.

Last edited by -TL- (2006-10-18 16:51:27)

2006-10-18 16:51:27

#11

Leprechaun56: Proud Infantry Whore; +31|6581|U.S.A

i think it wanted a number. Do i remember correctly?

2006-10-18 16:53:59

#12

Flaming_Maniac: prince of insufficient light; +2,490|6703|67.222.138.85

lmao, I had that exact same problem.

the answer is zero

If something times x say to the power of 2 equals the x to the power of a+1 or in this case now 2, then ak^2=a^3. This means that a=k, so a-k = 0.

Plug in numbers for a and x helps.

k(3)^1 = (3)^2
k(3)=9
k=3=x

I'm pretty sure I did all the math part right, I might be able to help on other problems.

Last edited by Flaming_Maniac (2006-10-18 16:57:13)

2006-10-18 16:54:21

#13

vpyroman: Aeon Supreme commander; +16|6613|UCF

well kax = x^(a+1) => kax = x*x^a,
ka = x^a <- Right here is where I get stuck, Algebra 2 was a long time ago, Calc 2 is a far cry from this
(ka)^(1/a) = x
k-(ka)^(1/a) = Integer

2006-10-18 16:56:03

#14

Varegg: Support fanatic :-); +2,206|6807|Nårvei

A tad since i enjoyed math but since you do have x and a at both sides can`t you determine the value of those two yourself ?

Say x = 2
And a = 3

a+1 = 4

2⁴ = 32

k+2+3 = 32

32 - 5 = k

k = 27

27 - 3 = 24 <-----

Say x = 4
And a = 5

a+1 = 6

4⁶ = 256

k+4+5 = 256

256 - 9 = k

k = 247

247 - 5 = 242 <----

So i conclude that since 24 and 242 have much the same numbers k - a can be whatever you want it to be

Last edited by Varegg (2006-10-18 16:58:44)

Wait behind the line ..............................................................

2006-10-18 16:58:14

#15

Goven: /̵͇̿̿/'̿'̿ ̿; +125|6477|Purdue

Hmm. I decided to use k=4, a=1, and x=4 and worked out the problem.

https://img247.imageshack.us/img247/3547/mathnc0.png

So basically, if k and x are the same integer, k-a could be anywhere from 0-∞

2006-10-18 16:58:36

#16

>LOD<Dougalachi: Teh_Complainer; +85|6552|An Hour North of Indy

lol...i took them as well and that WAS a hard question. Didn you answer the one like this?:

the mean of a,b,c,d, and e is 180.

if the mean of a,b,c, and d is 140, what is E?

2006-10-18 16:59:56

#17

Flaming_Maniac: prince of insufficient light; +2,490|6703|67.222.138.85

You silly people, way overthinking the problem

2006-10-18 17:00:47

#18

Goven: /̵͇̿̿/'̿'̿ ̿; +125|6477|Purdue

>LOD<Dougalachi wrote:
lol...i took them as well and that WAS a hard question. Didn you answer the one like this?:

the mean of a,b,c,d, and e is 180.

if the mean of a,b,c, and d is 140, what is E?

Yea, it was the mean of a,b,c,d,e = 120
the mean of a,b,c,d = 80
but it was like t,u,v,w, and x
I just guessed and put 40 >.> I lost 1/4 of a point xD
But it's cool though. When we got our scores in December, we get the test books too.

Last edited by Goven (2006-10-18 17:01:32)

2006-10-18 17:01:16

#19

Flaming_Maniac: prince of insufficient light; +2,490|6703|67.222.138.85

>LOD<Dougalachi wrote:
lol...i took them as well and that WAS a hard question. Didn you answer the one like this?:

the mean of a,b,c,d, and e is 180.

if the mean of a,b,c, and d is 140, what is E?

Yeah, you divide 180 by 5 and 140 by 4, then find the difference, which = e. I don't think those were the right numbers though.

Edit: goven set up the problem right, but what I said is still how you solve it.

Last edited by Flaming_Maniac (2006-10-18 17:01:56)

2006-10-18 17:02:36

#20

>LOD<Dougalachi: Teh_Complainer; +85|6552|An Hour North of Indy

that stuff was pretty hard at the end...shouldve studied more. oh well, its only PSATs..Btw, where did you take it at (what state)?

2006-10-18 17:02:59

#21

Flaming_Maniac: prince of insufficient light; +2,490|6703|67.222.138.85

Goven wrote:
Hmm. I decided to use k=4, a=1, and x=4 and worked out the problem.
http://img247.imageshack.us/img247/3547/mathnc0.png
So basically, if k and x are the same integer, k-a could be anywhere from 0-∞

Except the problem didn't ask what k equals, it asked what k-x equals, and since they are the same number, the answer is zero.

Edit: I'm from Texas

Last edited by Flaming_Maniac (2006-10-18 17:03:17)

2006-10-18 17:06:18

#22

Goven: /̵͇̿̿/'̿'̿ ̿; +125|6477|Purdue

Indiana here, and I could've sworn that it said k-a

2006-10-18 17:08:35

#23

>LOD<Dougalachi: Teh_Complainer; +85|6552|An Hour North of Indy

lol...indiana as well for me.

2006-10-18 17:09:00

#24

Goven: /̵͇̿̿/'̿'̿ ̿; +125|6477|Purdue

>LOD<Dougalachi wrote:
lol...indiana as well for me.

What area?
Just read your title :b
about 45 minutes southeast of Indy

Last edited by Goven (2006-10-18 17:10:13)

2006-10-18 17:10:08

#25

Flaming_Maniac: prince of insufficient light; +2,490|6703|67.222.138.85

Goven wrote:
Indiana here, and I could've sworn that it said k-a

No, and that doesn't make any sense since those numbers could be anything.

BF2S Forums Math question.

-TL- wrote:

Varegg wrote:

=Robin-Hood= wrote:

evilcartman99 wrote:

-TL- wrote:

>LOD<Dougalachi wrote:

>LOD<Dougalachi wrote:

Goven wrote:

>LOD<Dougalachi wrote:

Goven wrote:

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