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# 5^12+5^13= A. 5^25 B. 10^25 C. 6(5^12) D. 10^12+5 E. 2

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Senior Manager
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5^12+5^13= A. 5^25 B. 10^25 C. 6(5^12) D. 10^12+5 E. 2 [#permalink]  27 Jul 2007, 20:45
5^12+5^13=

A. 5^25
B. 10^25
C. 6(5^12)
D. 10^12+5
E. 2 (5^12)+5

Can someone explain how to get answer?
These number properties kill me [/code]
Director
Joined: 26 Feb 2006
Posts: 905
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Re: Hard Problem..for me:) [#permalink]  27 Jul 2007, 20:49
jamesrwrightiii wrote:
5^12+5^13=

A. 5^25
B. 10^25
C. 6(5^12)
D. 10^12+5
E. 2 (5^12)+5

Can someone explain how to get answer?
These number properties kill me [/code]

just take out the common factor:
= 5^12+5^13
= 5^12 (1 + 5)
= 5^12 (6)
Senior Manager
Joined: 02 May 2004
Posts: 313
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Re: Hard Problem..for me:) [#permalink]  28 Jul 2007, 12:59
Himalayan wrote:
jamesrwrightiii wrote:
5^12+5^13=

A. 5^25
B. 10^25
C. 6(5^12)
D. 10^12+5
E. 2 (5^12)+5

Can someone explain how to get answer?
These number properties kill me [/code]

just take out the common factor:
= 5^12+5^13
= 5^12 (1 + 5)
= 5^12 (6)

That is the problem..I don't understand how to factor these types of equations..Manhattan GMAT doesn't even give a good explanation:(
Any web pages that explain the theory behind this?
I had up to Calc 2 and never encounted number properties
VP
Joined: 29 Apr 2003
Posts: 1405
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Re: Hard Problem..for me:) [#permalink]  28 Jul 2007, 13:06
jamesrwrightiii wrote:
Himalayan wrote:
jamesrwrightiii wrote:
5^12+5^13=

A. 5^25
B. 10^25
C. 6(5^12)
D. 10^12+5
E. 2 (5^12)+5

Can someone explain how to get answer?
These number properties kill me [/code]

just take out the common factor:
= 5^12+5^13
= 5^12 (1 + 5)
= 5^12 (6)

That is the problem..I don't understand how to factor these types of equations..Manhattan GMAT doesn't even give a good explanation:(
Any web pages that explain the theory behind this?
I had up to Calc 2 and never encounted number properties

Well I guess it come with practice. After a while when u c the problem, you can identify what to do. Like the fact that there were 5^X in both the terms in the given expression should have been a hint to take the term with the lower value out.

Just practice, you will be fine.
Senior Manager
Joined: 02 May 2004
Posts: 313
Followers: 1

Kudos [?]: 6 [0], given: 0

Are there any rules/equations that I am missing?
VP
Joined: 29 Apr 2003
Posts: 1405
Followers: 2

Kudos [?]: 18 [0], given: 0

jamesrwrightiii wrote:
Are there any rules/equations that I am missing?

ab + bc = b(a+c)

its a simple simplification rule.
Manager
Joined: 18 Jun 2007
Posts: 69
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Why cant I add an attachement ?/???????
Senior Manager
Joined: 14 Jun 2007
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Re: Hard Problem..for me:) [#permalink]  28 Jul 2007, 19:05
jamesrwrightiii wrote:
Himalayan wrote:
jamesrwrightiii wrote:
5^12+5^13=

A. 5^25
B. 10^25
C. 6(5^12)
D. 10^12+5
E. 2 (5^12)+5

Can someone explain how to get answer?
These number properties kill me [/code]

just take out the common factor:
= 5^12+5^13
= 5^12 (1 + 5)
= 5^12 (6)

That is the problem..I don't understand how to factor these types of equations..Manhattan GMAT doesn't even give a good explanation:(
Any web pages that explain the theory behind this?
I had up to Calc 2 and never encounted number properties

If you went up to Calc II you encountered factoring out numbers! It comes up frequently when you have to factor something outside of an integral for the power rule. I am a bit rusty, but I hope this helps... it is the same concept, and should be familiar to you if you went up to Calc II (I learned this in a Applied Business Calc class):

S= integration sign
by the power rule:

S[u(x)]^n*u'(x)dx

Su^ndu = (u^(n+1))/(n+1)+C if u=u(x) du = u'dx n =/= 1

take the integral:
S sqrt((x^3)-4) * (5x^2)dx

u= x^3=4
u '= 3x^2
n = 1/2

since the derivative of u = 3x^2 we can see that inegral is not in the correct form because we have 5x^2 so factor out 5/3

(5/3)S((x^3)-4)^1/2*3x^2dx

which gives you the indefinitte integral for the family of functions defined by:

(10/9)((x^3)-4)^(3/2)+C

phew..
CEO
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Re: Hard Problem..for me:) [#permalink]  28 Jul 2007, 21:51
jamesrwrightiii wrote:
5^12+5^13=

A. 5^25
B. 10^25
C. 6(5^12)
D. 10^12+5
E. 2 (5^12)+5

Can someone explain how to get answer?
These number properties kill me [/code]

jamesrwrightiii wrote:
5^12+5^13=

A. 5^25
B. 10^25
C. 6(5^12)
D. 10^12+5
E. 2 (5^12)+5

Can someone explain how to get answer?
These number properties kill me [/code]

5^12+5^13... hmmmm ok first advice don't try and figure out what 5^12 is.

Just take out factors. 5^12+5^13-> 5^12(1+5) 5^12(6)

Now if you were like me and didn't realize to do this in 2minutes. You could have taken the alternative approach through process of elmination and by knowing a few exponent properties.

A is clearly out you can only add the xponents w/ same base AND when they are multiplied by each other. ie 5^2*5^3=5^5

B is out for the same reason. Also it makes the mistake of adding the two 5's as well

C... couldn't say no to this, although I wanted to because of the 6.

D. 10^12+5? This choice is just silly. While it factored out the 5, to get the bases and exponents equal, you can't just add them. Same reasoning as I mentioned in A.

E. E was a bit tricky b/c it kinda made sense. however, 2(5^12)+5 is not the same as 5^12+5^13. Just try it out w/ basic numbers. (3^2)2+3 does not equal 3^2+3^3... also this suggests that somehow the original stmnt had 5^12*5^12. Clearly not the answer.

C was the only one left so I picked it. Then bout 10sec later i realized I could do what I first wrote out... doh!
Re: Hard Problem..for me:)   [#permalink] 28 Jul 2007, 21:51
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