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jimmyjamesdonkey
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Very Very Basic Math Question 11 Jun 2007, 07:04
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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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Fig
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Very Very Basic Math Question 11 Jun 2007, 07:19
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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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KillerSquirrel
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Very Very Basic Math Question Updated on: 11 Jun 2007, 07:28
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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(r^3)(r^3)= r*r*r*r*r*r = r^6 -

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

:?
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jimmyjamesdonkey
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Very Very Basic Math Question 11 Jun 2007, 07:27
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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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KillerSquirrel
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Very Very Basic Math Question 11 Jun 2007, 07:33
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jimmyjamesdonkey
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.
Show more


{ ( (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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jimmyjamesdonkey
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Very Very Basic Math Question 11 Jun 2007, 07:38
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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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Very Very Basic Math Question 11 Jun 2007, 07:42
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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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KillerSquirrel
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Very Very Basic Math Question 11 Jun 2007, 07:44
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jimmyjamesdonkey
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 :)
Show more


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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Very Very Basic Math Question 11 Jun 2007, 07:45
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jimmyjamesdonkey
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 :)
Show more


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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Very Very Basic Math Question 11 Jun 2007, 07:50
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jimmyjamesdonkey
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.
Show more


{ ( (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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KillerSquirrel
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Very Very Basic Math Question 11 Jun 2007, 07:52
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Fig
jimmyjamesdonkey
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
Show more


good explanation - Fig

:)
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Very Very Basic Math Question 11 Jun 2007, 07:59
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KillerSquirrel
Fig
jimmyjamesdonkey
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

:)
Show more


Thanks :)



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Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Quantitative Questions Forum
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.
Thank you for understanding, and happy exploring!