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Re: If a and b are different nonzero integers, what is the value of b? (1 [#permalink]

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04 May 2015, 09:43

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This post received KUDOS

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According to statement 1: a^b = ab a^(b-1) = b So (a,b) = (2,2) or (any integer, 1) But since a and b are different, the first option can be discarded. So b can only be 1. SUFFICIENT

According to statement 2: a^b - a^(b-1) = 2 (a^(b-1))*(a-1) = 2 Either a^(b-1) = 2 and a-1 = 1, i.e. (a,b) = (2,2) or a^(b-1) = 1 and a-1 = 2 i.e. (a,b) = (3,1) or a^(b-1) = -2 and a-1 = -1 which is not possible as a is non-zero or a^(b-1) = -1 and a - 1 = -2 i.e. (a,b) = (-1,2)

The first option is not possible as a and b are different but second and fourth are possible, which gives us different values of b. NOT SUFFICIENT

A is the right choice here.

Please note that if you remove the condition that a and b are 'different,' then two values of b are possible through statement (1) as well and E becomes the right choice.

Last edited by PrepTap on 05 May 2015, 10:27, edited 1 time in total.

(1) SUFFICIENT: If b=1, then a can be any number whatsoever, because a^1 and a(1) will always be the same number. Therefore the issue is to determine whether b can be any number other than 1.

The given equation can be rephrased. Divide both sides by a, giving a^{b-1}=b.

First, consider the case in which b is positive.

If b=2, then this equation reduces to a^1=2, so that a=2; however, this case can be disregarded, because a and b are not allowed to be the same.

If b=3 or more, the equation is a^2=3, a^3=4, a^4=5, etc. None of these equations have integer solutions.

What if b is negative?

If b=-1, then a^{-2}=-1, which is impossible! Something raised to an even power cannot be negative.

If b=-2 or less, then we have a^{-3}=-2, a^{-4}=-3, etc. These equations are all impossible. If the power is even, then the expression can’t be negative. If the power is odd, the resulting value of a will be a fraction.

This rules out every case except b=1. The statement is sufficient.

(2) NOT SUFFICIENT: To find cases that simplify this equation, it is helpful to factor it. Pull an a^{b-1} out of both terms on the right side: a^{b-1}(a-1)=2. (If you’re not sure why what’s the right factorization, write out the steps yourself; that’ll be a great test of your knowledge of exponent rules).

Now test some cases:

If a=3 and b=1, then the expression is 3^0(3-1)=1(2)=2.

If a=1 and b=-2, then the expression is -1^{-3}(-1-1)=(-1)(-2)=2.

There are at least two possibilities for b, so this statement can’t be sufficient. It turns out that if a=-1 and b= any even number, you’ll get 2 as the answer (not sure why? test some even numbers to see whether you can figure out the rule). Therefore, b can be 1 or any even number.

Re: If a and b are different nonzero integers, what is the value of b? (1 [#permalink]

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11 May 2015, 07:49

Bunuel wrote:

If a and b are different nonzero integers, what is the value of b?

(1) \(a^b = ab\)

(2) \(a^b - a^{b - 1} = 2\)

Kudos for a correct solution.

a≠b and a ≠0

1) ab=ab a(a^(b-1) -1 ) = 0

either a =0 or ab-1 =1 if and b are intergers only way possible for this is for b = 1 2) a^b−a^(b−1)=2 a^b ( 1 – 1/a) = 2 a^( b-1) * ( a- 1) = 2 ( a = 3 and b = 1 )and (a =2 and b = 2) both will satisfy this equation.

Re: If a and b are different nonzero integers, what is the value of b? (1 [#permalink]

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23 Sep 2017, 08:23

Hello from the GMAT Club BumpBot!

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