How many factors does the integer X have? 1. X^(x+3) =

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How many factors does the integer X have? 1. X^(x+3) = [#permalink]

04 Oct 2010, 06:16

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How many factors does the integer X have?

1. X^(x+3) = (2x)^(x-1)
2. |3x-7| = 2x+2

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Re: DS: Factors of integer X [#permalink]

04 Oct 2010, 06:30

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eladshush wrote:

How many factors does the integer X have?

1. X^(x+3) = (2x)^(x-1)
2. |3x-7| = 2x+2

(1) x^{(x+3)}=(2x)^{(x-1)} --> x^{(x+3)}=2^{(x-1)}*x^{(x-1)} --> x^{(x+3-x+1)}=2^{(x-1)} --> x^4=2^{(x-1)} --> x=1. Sufficient.

(2) |3x-7| = 2x+2 --> x=1 or x=9. Not sufficient.

Answer: A.
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Re: DS: Factors of integer X [#permalink]

04 Oct 2010, 09:12

eladshush wrote:

How many factors does the integer X have?

1. X^(x+3) = (2x)^(x-1)
2. |3x-7| = 2x+2

(1) x^{x+3}=(2x)^{x-1}
x^x * x^3 = 2^{x-1} * x^{-1} * x^x
Assuming x is not 0, we can cancel x^x out
x^4 = 2^{x-1}
The only solution to this is x=1.
1 has only one factor
Sufficient

(2) x>=7/3 then 3x-7=2x+2, x=9
x<7/3 then 3x-7=-2x-2 , x=1
x can have 1 factor or 3 factors
Insufficient

Answer is (A)
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Re: How many factors does the integer X have? 1. X^(x+3) = [#permalink]

14 Aug 2012, 00:47

how can the answer be A wont x = 0 satify equation no 1 ?...expert plz reply !!

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Re: How many factors does the integer X have? 1. X^(x+3) = [#permalink]

14 Aug 2012, 01:24

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aditya8062 wrote:

how can the answer be A wont x = 0 satify equation no 1 ?...expert plz reply !!

If x=0 in (1) then the left hand side of the equation becomes 0^{-1}=\frac{1}{0}=undefined, (while the right hand side becomes 0^3=0). Remember you cannot raise zero to a negative power.

Hope it helps.
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Re: How many factors does the integer X have? 1. X^(x+3) = [#permalink]

15 Aug 2012, 08:44

eladshush wrote:

How many factors does the integer X have?

1. X^(x+3) = (2x)^(x-1)
2. |3x-7| = 2x+2

here, wat is meant by "How many factors" ?

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Re: How many factors does the integer X have? 1. X^(x+3) = [#permalink]

15 Aug 2012, 08:47

mehulsayani wrote:

eladshush wrote:

How many factors does the integer X have?

1. X^(x+3) = (2x)^(x-1)
2. |3x-7| = 2x+2

here, wat is meant by "How many factors" ?

The question asks about (positive) divisors of some integer. For example 6 has 4 divisors (factors): 1, 2, 3, and 6.

For more check Number Theory chapter of Math Book: math-number-theory-88376.html

Hope it helps.
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Re: DS: Factors of integer X [#permalink]

17 Aug 2012, 05:58

Bunuel wrote:

eladshush wrote:

How many factors does the integer X have?

1. X^(x+3) = (2x)^(x-1)
2. |3x-7| = 2x+2

(1) x^{(x+3)}=(2x)^{(x-1)} --> x^{(x+3)}=2^{(x-1)}*x^{(x-1)} --> x^{(x+3-x+1)}=2^{(x-1)} --> x^4=2^{(x-1)} --> x=1. Sufficient.

(2) |3x-7| = 2x+2 --> x=1 or x=9. Not sufficient.

Answer: A.

x^4=2^{(x-1)} --> x=1---------How do we arrive at this. How to solve these????? just by testing values vaguely or is there any particular mathematics we can do with this?????

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Re: DS: Factors of integer X [#permalink]

17 Aug 2012, 06:26

GMATPASSION wrote:

Bunuel wrote:

eladshush wrote:

How many factors does the integer X have?

1. X^(x+3) = (2x)^(x-1)
2. |3x-7| = 2x+2

(1) x^{(x+3)}=(2x)^{(x-1)} --> x^{(x+3)}=2^{(x-1)}*x^{(x-1)} --> x^{(x+3-x+1)}=2^{(x-1)} --> x^4=2^{(x-1)} --> x=1. Sufficient.

(2) |3x-7| = 2x+2 --> x=1 or x=9. Not sufficient.

Answer: A.

x^4=2^{(x-1)} --> x=1---------How do we arrive at this. How to solve these????? just by testing values vaguely or is there any particular mathematics we can do with this?????

Number plugging is the best way to solve x^4=2^{(x-1)}:

x=0 does not satisfy the equation (check this: how-many-factors-does-the-integer-x-have-1-x-x-102180.html#p1112836).
x=1 satisfies the equation.

Now, if x>1 (2, 3, ...), then the left hand side of the equation is always greater than the right hand side of the equation: 2^4=16>2^1=2, 3^4=81>2^2=4, ... (as you can see LHS increases "faster" than RHS).

Hence, x=1 is the only positive integer solution of x^4=2^{(x-1)}.

Hope it helps.
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Re: DS: Factors of integer X [#permalink] 17 Aug 2012, 06:26

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How many factors does the integer X have? 1. X^(x+3) =

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