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If a and b are different nonzero integers, what is the value of b? (1

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

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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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New post 04 May 2015, 07:33
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Bunuel wrote:
If a and b are different nonzero integers, what is the value of b?

(1) ab = ab

(2) ab – ab – 1 = 2


Kudos for a correct solution.


Is this question correct?
Did you mean to say:
1. ab = a/b
2. ab - a/b -1 = 2

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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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New post 04 May 2015, 07:38
PrepTap wrote:
Bunuel wrote:
If a and b are different nonzero integers, what is the value of b?

(1) ab = ab

(2) ab – ab – 1 = 2


Kudos for a correct solution.


Is this question correct?
Did you mean to say:
1. ab = a/b
2. ab - a/b -1 = 2


Edited. Thank you for noticing this.
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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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New post 04 May 2015, 09:18
St 1 says

a^b = ab

1st case a=2 and b=1, works fine 2^1 = 2*1 => 2=2

what if 2^2 = 2*2 => 4=4,

not sufficient.

st#2 says

a^b−a^b−1=2

Again picking numbers a=3 and b=1 => 3 - 3^0 => 3-1 =2, works fine.

but if a=2 and b=2 => 2^2 - 2^(2-1) => 4 - 2 =2 , works fine.

not sufficient.

combining both statements same result got two values of b 1 &2.

Answer E

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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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New post 04 May 2015, 09:43
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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.

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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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New post 05 May 2015, 02:03
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Got tricked initially (did the same mistake), good question !! (need to take note that a,b are different non-zero integers)

Q : If a and b are different and nonzero integers , what is the value of b?
    make sure to use all the information provided in question stem, question stem tells us that

    (i) \(a\neq{0}\) or \(b\neq{0}\)
    (ii) \(a\neq{b}\)

now to the statements
stmt 1: \(a^{b}\) = ab

    (a,b) = (1,1), (2,2) , (3,1) , (4,1)

    1^1 = 1*1 --> ignore this a,b are different non-zero
    2^2 = 2*2 --> ignore this a,b are different non-zero
    3^1 = 3*1
    4^1 = 4*1...

b is always 1---> b =1 got one solution , so Sufficient

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

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

    this equation works for values as above (a,b) = (2,2) , (3,1) (refer below with inserted values)

    \(2^{2}\) (1-\(\frac{1}{2}\)) => 4 (\(\frac{1}{2}\)) = 2

    (2,2) this works But, ignore this a,b are different non-zero

    \(3^{1}\) (1-\(\frac{1}{3}\)) => 3 (\(\frac{2}{3}\)) = 2

    (a,b) = (3,1) works

    [Edited] got one more solution

    \((-1)^{2}\) (1-\(\frac{1}{(-1)}\)) => 1 (1+1) = 2
    (a,b) = (-1,2) did works ---> b got two solution , Not Sufficient


so both stmt 1 & stmt2 are Sufficient:

Ans : A
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Last edited by UJs on 07 May 2015, 01:26, edited 6 times in total.

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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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New post 05 May 2015, 06:41
UjjwalS wrote:
required value of b ?

Got tricked initially (did the same mistake ), good question ( need to take note that a,b are different non-zero integers)

stmt 1: a^b = ab

    (a,b) = (1,1), (2,2) , (3,1) , (4,1)

    1^1 = 1*1 --> ignore this a,b are different non-zero
    2^2 = 2*2 --> ignore this a,b are different non-zero
    3^1 = 3*1
    4^1 = 4*1...

b is always 1---> b =1 got one solution , so Sufficient

stmt 2: a^b - a^b-1= 2

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

    this equation works for values as above (a,b) = (2,2) , (3,1) (refer below with inserted values)

    2^2 (1-\(\frac{1}{2}\)) => 4 (\(\frac{1}{2}\)) = 2

    (2,2) this works But, ignore this a,b are different non-zero

    3^1 (1-\(\frac{1}{3}\)) => 3 (\(\frac{2}{3}\)) = 2

    (a,b) = (3,1) works ---> b got one solution , so Sufficient

so both stmt 1 & stmt2 are Sufficient:
Ans : D



What about a=2 and b=2

2^2 = 4

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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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New post 05 May 2015, 09:29
Quote:
What about a=2 and b=2

2^2 = 4

viksingh15
it is tricky, make sure to use all the information provided in question stem.I understand the confusion.

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

it tells us that

    1> \(a\neq{0}\) or \(b\neq{0}\)

    2> \(a\neq{b}\)

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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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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.


MANHATTAN GMAT OFFICIAL SOLUTION:

(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.

The correct answer is A.
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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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New post 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.

Answer A

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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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