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Very Very Basic Math Question

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jimmyjamesdonkey
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Very Very Basic Math Question [#permalink] New post   11 Jun 2007, 07:04
Hi All,

I have a very basic math question I was hoping all could answer.

According to the ManhattanGMAT Number Properties guide the following is true:

(3^3)(5^3) = 15^3

Basically, when multiplying expressions with the same exp, multiple the bases first. Ok, so please explain this:

(r^3)(r^3)=?

That should equal r^6, but according to ManhattanGMAT rules wouldn't it be r^3 (incorrect)?

Thanks.



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Re: Very Very Basic Math Question [#permalink] New post   11 Jun 2007, 07:19
The reasonning is correct:

o (3^3)(5^3)
= (5*3)^3
= (15)^3

o (r^3)(r^3)
= (r*r)^3
= (r^2)^3
= r^(2*3)
= r^6

Hope that helps :)
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Re: Very Very Basic Math Question [#permalink] New post   Updated on: 11 Jun 2007, 07:28
(r^3)(r^3)= r*r*r*r*r*r = r^6 -

I can't see how you can get r^3

:?

Originally posted by KillerSquirrel on 11 Jun 2007, 07:27.
Last edited by KillerSquirrel on 11 Jun 2007, 07:28, edited 1 time in total.
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jimmyjamesdonkey
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Re: Very Very Basic Math Question [#permalink] New post   11 Jun 2007, 07:27
Please explain this problem then:

{ ( (m^8)(p^7)(r^12) ) / ( (m^3)(r^9)(p) ) } * ((p^2)(r^3)(m^4))

I first divide the fraction and then multiple getting:
m^9 * p^8 * r^3 (r^3 since doing it this way we have like r terms and exp so I keep the exp)

OA is m^9 * p^8 * r^6, because they multiplied first and then divided. Please explain why we get different answers based on how the questions is solved.
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Re: Very Very Basic Math Question [#permalink] New post   11 Jun 2007, 07:33
jimmyjamesdonkey wrote:
Please explain this problem then:

{ ( (m^8)(p^7)(r^12) ) / ( (m^3)(r^9)(p) ) } * ((p^2)(r^3)(m^4))

I first divide the fraction and then multiple getting:
m^9 * p^8 * r^3 (r^3 since doing it this way we have like r terms and exp so I keep the exp)

OA is m^9 * p^8 * r^6, because they multiplied first and then divided. Please explain why we get different answers based on how the questions is solved.


{ ( (m^8)(p^7)(r^12) ) / ( (m^3)(r^9)(p) ) } * ((p^2)(r^3)(m^4)) =

(m^5)(p^6)(r^3) * (p^2)(r^3)(m^4)=

(m^9)(p^8)(r^6)

since m^2/m = m*m/m = m

and since r^3*r^2 = r*r*r*r*r = r^5

:-D
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Re: Very Very Basic Math Question [#permalink] New post   11 Jun 2007, 07:38
I guess there is two rules going on at the same time!

Rule 1: When multiplying expressions with the same base, ADD the exp first.
Rule 2: When multiplying expression with the same exp, multiply bases first.

In this case r3 * r3 we have BOTH the same base and same exponent! So I guess when that happens, we add the exponents (like the rule says "first", duh) :)

FYI...I knew how to do this with no prob BEFORE reading the prep book and confusing me :)
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Re: Very Very Basic Math Question [#permalink] New post   11 Jun 2007, 07:42
To answer this question, you may use the following algebric property:

x^y/x^m = x^y . x^-m = x^(y-m)

use this for simplifying the early experession between the square brackets, and then simply multiply the variables [ by summing the exponents ] to arrive at the correct OA answer you provided. If you need me to show you the arithmatic in details, I'll more than glad to do so.
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Re: Very Very Basic Math Question [#permalink] New post   11 Jun 2007, 07:44
jimmyjamesdonkey wrote:
I guess there is two rules going on at the same time!

Rule 1: When multiplying expressions with the same base, ADD the exp first.
Rule 2: When multiplying expression with the same exp, multiply bases first.

In this case r3 * r3 we have BOTH the same base and same exponent! So I guess when that happens, we add the exponents (like the rule says "first", duh) :)

FYI...I knew how to do this with no prob BEFORE reading the prep book and confusing me :)


If you get confuse - think about this:

r^3*r^3 = r*r*r*r*r*r = r^6 meaning - rule one

r^3*m^3 = r*r*r*m*m*m = (r*m)*(r*m)*(r*m) = (r*m)^3 meaning rule 2

:-D
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Re: Very Very Basic Math Question [#permalink] New post   11 Jun 2007, 07:45
jimmyjamesdonkey wrote:
I guess there is two rules going on at the same time!

Rule 1: When multiplying expressions with the same base, ADD the exp first.
Rule 2: When multiplying expression with the same exp, multiply bases first.

In this case r3 * r3 we have BOTH the same base and same exponent! So I guess when that happens, we add the exponents (like the rule says "first", duh) :)

FYI...I knew how to do this with no prob BEFORE reading the prep book and confusing me :)


The bold is right but u could have the different way to make it as the one used by MGMAT....

(r^2)^3 = r^(2*3) = r^(3*2) = (r^3)^2 = r^6

It's important to keep in mind that both exist :)... It's sometimes tested ;)
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Re: Very Very Basic Math Question [#permalink] New post   11 Jun 2007, 07:50
jimmyjamesdonkey wrote:
Please explain this problem then:

{ ( (m^8)(p^7)(r^12) ) / ( (m^3)(r^9)(p) ) } * ((p^2)(r^3)(m^4))

I first divide the fraction and then multiple getting:
m^9 * p^8 * r^3 (r^3 since doing it this way we have like r terms and exp so I keep the exp)

OA is m^9 * p^8 * r^6, because they multiplied first and then divided. Please explain why we get different answers based on how the questions is solved.


{ ( (m^8)(p^7)(r^12) ) / ( (m^3)(r^9)(p) ) } * ((p^2)(r^3)(m^4))
= m^a * p^b * r^c

Where u count for a, b and c :
o plus an exponent if in the numerator
o minus an exponent if in the denominator

So,
o a = 8 - 3 + 4 = 9
o b = 7 - 1 + 2 = 8
o c = 12 - 9 + 3 = 6
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Re: Very Very Basic Math Question [#permalink] New post   11 Jun 2007, 07:52
Fig wrote:
jimmyjamesdonkey wrote:
Please explain this problem then:

{ ( (m^8)(p^7)(r^12) ) / ( (m^3)(r^9)(p) ) } * ((p^2)(r^3)(m^4))

I first divide the fraction and then multiple getting:
m^9 * p^8 * r^3 (r^3 since doing it this way we have like r terms and exp so I keep the exp)

OA is m^9 * p^8 * r^6, because they multiplied first and then divided. Please explain why we get different answers based on how the questions is solved.


{ ( (m^8)(p^7)(r^12) ) / ( (m^3)(r^9)(p) ) } * ((p^2)(r^3)(m^4))
= m^a * p^b * r^c

Where u count for a, b and c :
o plus an exponent if in the numerator
o minus an exponent if in the denominator

So,
o a = 8 - 3 + 4 = 9
o b = 7 - 1 + 2 = 8
o c = 12 - 9 + 3 = 6


good explanation - Fig

:)
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Re: Very Very Basic Math Question [#permalink] New post   11 Jun 2007, 07:59
KillerSquirrel wrote:
Fig wrote:
jimmyjamesdonkey wrote:
Please explain this problem then:

{ ( (m^8)(p^7)(r^12) ) / ( (m^3)(r^9)(p) ) } * ((p^2)(r^3)(m^4))

I first divide the fraction and then multiple getting:
m^9 * p^8 * r^3 (r^3 since doing it this way we have like r terms and exp so I keep the exp)

OA is m^9 * p^8 * r^6, because they multiplied first and then divided. Please explain why we get different answers based on how the questions is solved.


{ ( (m^8)(p^7)(r^12) ) / ( (m^3)(r^9)(p) ) } * ((p^2)(r^3)(m^4))
= m^a * p^b * r^c

Where u count for a, b and c :
o plus an exponent if in the numerator
o minus an exponent if in the denominator

So,
o a = 8 - 3 + 4 = 9
o b = 7 - 1 + 2 = 8
o c = 12 - 9 + 3 = 6


good explanation - Fig

:)


Thanks :)



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This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
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Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
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Re: Very Very Basic Math Question [#permalink]
11 Jun 2007, 07:59
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