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x is an integer and x raised to any odd integer is greater

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x is an integer and x raised to any odd integer is greater  [#permalink]

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x is an integer and x raised to any odd integer is greater than zero; is w - z greater than 5 times the quantity 7^(x-1)-5^x?

(1) z < 25 and w=7^x
(2) x = 4

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Originally posted by ykaiim on 08 Jun 2010, 01:14.
Last edited by Bunuel on 20 May 2012, 11:57, edited 3 times in total.
Edited the question and added the OA
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Re: DS problem  [#permalink]

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New post 08 Jun 2010, 04:58
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ykaiim wrote:
x is an integer and x raised to any odd integer is greater than zero; is w - z greater than 5 times the quantity \(7^x^-^1 - 5^x\)?

1) z < 25 and w = 7x
2) x = 4


Note: The above expression is 7^(x-1) -5^x.


I guess statement (1) should be: \(z<25\) and \(w=7^x\).

"x is an integer and x raised to any odd integer is greater than zero" means \(x=integer>0\).
Q: is \(w-z>5(7^{x-1}-5^x)\)?

(1) \(z<25\) and \(w=7^x\) --> is \(7^x-z>5(7^{x-1}-5^x)\)? --> is \(7^x-z>5*7^{x-1}-5^{x+1}\)? --> is \(7^x-5*7^{x-1}+5^{x+1}>z\)? --> as the lowest value of \(x\) is 1, then the lowest value of LHS is when \(x=1\): \(LHS_{min}=7^x-5*7^{x-1}+5^{x+1}=7-5+25=27\) --> so the lowest value of LHS is 27, which is more than \(z\) as \(z<25\) --> hence \(7^x-5*7^{x-1}+5^{x+1}>z\) is true. Sufficient.

(2) \(x = 4\). No info about \(w\) and \(z\). Not sufficient.

Answer: A.
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Re: Is w - z greater than 5 times the quantity 7^x-1 - 5^x  [#permalink]

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New post 20 May 2012, 09:33
Could you please reply with the detailed explanation for the above question.
i am confused as to how just Statement I is sufficient to answer the question(A), without knowing the value of X?
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Re: Is w - z greater than 5 times the quantity 7^x-1 - 5^x  [#permalink]

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New post 20 May 2012, 09:52
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Hello, let me try and explain.

paraphrasing the questions gives us, is w-z > 5.7^(x-1) - 5^(x+1). Further, x is a positive integer as an odd power of x is greater than 0.

taking statement 1, you have the question as 7^x-z > 5.7^(x-1) - 5^(x+1)

Now, for all cases where x is a positive integer, 7^x would be greater than 5.7^(x-1). Next, 5^(x+1) would always be >= 25 as the least value of x can be 1.

So, 7^x > 5.7^x-1
and z < 5^(x+1). Reversing this inequality and adding both the inequalities, we get:

7^x - z > 5.7^x-1 - 5^(x+1),

Hence answer A.
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Re: Is w - z greater than 5 times the quantity 7^x-1 - 5^x  [#permalink]

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New post 20 May 2012, 10:21
the question is not simple but if you think a bit abstract and test number you can arrive to the solution.

The FIRST important thing, breaking the problem, is to understand tha our X is positive. Infact, if a number raised to power of 3 is >0 that means X itself is positive because the odd powr maintain the original sign of the number, so X must be positive.

At this point using the exponent rules we have w - z > 5*7^x-1 - 5^x+1 ----> 7^x - 24 (from stem x<25) > 5*7^x-1 - 5^x+1

testing number 1 ------> 7 -24 > 1 - 25 ---> - 17 > -24 this is TRUE. also if you test some value < 25 positive but also negative the result is the same.

1) sufficient

x = 4 -------> w - z > 5*7^3 - 5^5..........at this point you know nothing about w and z.

2) insufficient

A is the answer.
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Re: Is w - z greater than 5 times the quantity 7^x-1 - 5^x  [#permalink]

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New post 03 Jun 2012, 01:27
Bunuel wrote:
carcass wrote:
x is an integer and x raised to any odd integer is greater than zero; is w - z greater than 5 times the quantity 7^x-1 - 5^x?

1 z < 25 and w = 7^x
2 x = 4


Merging similar topics. Please ask if anything remains unclear.


why can't x be zero apart from even ? any odd number raised to the power zero is 1 which is an interger greater than zero. I think x is given as an interger and not only positive integer. if you work with x = 0 , statement 1 is not sufficient ?
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Re: Is w - z greater than 5 times the quantity 7^x-1 - 5^x  [#permalink]

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New post 03 Jun 2012, 02:38
Kaps07 wrote:
Bunuel wrote:
carcass wrote:
x is an integer and x raised to any odd integer is greater than zero; is w - z greater than 5 times the quantity 7^x-1 - 5^x?

1 z < 25 and w = 7^x
2 x = 4


Merging similar topics. Please ask if anything remains unclear.


why can't x be zero apart from even ? any odd number raised to the power zero is 1 which is an interger greater than zero. I think x is given as an interger and not only positive integer. if you work with x = 0 , statement 1 is not sufficient ?


We are told that integer x raised to to any odd integer is greater than zero: \(x^{odd}>0\). Now, if \(x=0\) then \(x^{odd}=0^{odd}=0\) (for odd>0), which violates given condition, so \(x\) cannot be zero.

Hope it's clear.
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Re: Is w - z greater than 5 times the quantity 7^x-1 - 5^x  [#permalink]

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New post 03 Jun 2012, 03:02
Thanks, I misunderstood the question. I thought it was odd int raised to the power x. It is clear now.
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Re: DS problem  [#permalink]

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New post 06 Sep 2012, 04:59
Bunuel wrote:
ykaiim wrote:
x is an integer and x raised to any odd integer is greater than zero; is w - z greater than 5 times the quantity \(7^x^-^1 - 5^x\)?

1) z < 25 and w = 7x
2) x = 4


Note: The above expression is 7^(x-1) -5^x.


I guess statement (1) should be: \(z<25\) and \(w=7^x\).

"x is an integer and x raised to any odd integer is greater than zero" means \(x=integer>0\).
Q: is \(w-z>5(7^{x-1}-5^x)\)?

(1) \(z<25\) and \(w=7^x\) --> is \(7^x-z>5(7^{x-1}-5^x)\)? --> is \(7^x-z>5*7^{x-1}-5^{x+1}\)? --> is \(7^x-5*7^{x-1}+5^{x+1}>z\)? --> as the lowest value of \(x\) is 1, then the lowest value of LHS is when \(x=1\): \(LHS_{min}=7^x-5*7^{x-1}+5^{x+1}=7-5+25=27\) --> so the lowest value of LHS is 27, which is more than \(z\) as \(z<25\) --> hence \(7^x-5*7^{x-1}+5^{x+1}>z\) is true. Sufficient.

(2) \(x = 4\). No info about \(w\) and \(z\). Not sufficient.

Answer: A.


I don't understand the function of statement z<25. If lowest value of LHS is 27 which is more than z, then it doesn't fulfill the condition given?
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Re: DS problem  [#permalink]

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New post 02 May 2013, 12:45
Bunuel wrote:
ykaiim wrote:
x is an integer and x raised to any odd integer is greater than zero; is w - z greater than 5 times the quantity \(7^x^-^1 - 5^x\)?

1) z < 25 and w = 7x
2) x = 4


Note: The above expression is 7^(x-1) -5^x.


I guess statement (1) should be: \(z<25\) and \(w=7^x\).

"x is an integer and x raised to any odd integer is greater than zero" means \(x=integer>0\).
Q: is \(w-z>5(7^{x-1}-5^x)\)?

(1) \(z<25\) and \(w=7^x\) --> is \(7^x-z>5(7^{x-1}-5^x)\)? --> is \(7^x-z>5*7^{x-1}-5^{x+1}\)? --> is \(7^x-5*7^{x-1}+5^{x+1}>z\)? --> as the lowest value of \(x\) is 1, then the lowest value of LHS is when \(x=1\): \(LHS_{min}=7^x-5*7^{x-1}+5^{x+1}=7-5+25=27\) --> so the lowest value of LHS is 27, which is more than \(z\) as \(z<25\) --> hence \(7^x-5*7^{x-1}+5^{x+1}>z\) is true. Sufficient.

(2) \(x = 4\). No info about \(w\) and \(z\). Not sufficient.

Answer: A.


Please explain why the lowest value of X is 1.
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Re: DS problem  [#permalink]

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New post 03 May 2013, 04:03
Rajkiranmareedu wrote:
Bunuel wrote:
ykaiim wrote:
x is an integer and x raised to any odd integer is greater than zero; is w - z greater than 5 times the quantity \(7^x^-^1 - 5^x\)?

1) z < 25 and w = 7x
2) x = 4


Note: The above expression is 7^(x-1) -5^x.


I guess statement (1) should be: \(z<25\) and \(w=7^x\).

"x is an integer and x raised to any odd integer is greater than zero" means \(x=integer>0\).
Q: is \(w-z>5(7^{x-1}-5^x)\)?

(1) \(z<25\) and \(w=7^x\) --> is \(7^x-z>5(7^{x-1}-5^x)\)? --> is \(7^x-z>5*7^{x-1}-5^{x+1}\)? --> is \(7^x-5*7^{x-1}+5^{x+1}>z\)? --> as the lowest value of \(x\) is 1, then the lowest value of LHS is when \(x=1\): \(LHS_{min}=7^x-5*7^{x-1}+5^{x+1}=7-5+25=27\) --> so the lowest value of LHS is 27, which is more than \(z\) as \(z<25\) --> hence \(7^x-5*7^{x-1}+5^{x+1}>z\) is true. Sufficient.

(2) \(x = 4\). No info about \(w\) and \(z\). Not sufficient.

Answer: A.


Please explain why the lowest value of X is 1.


See the highlighted part: x is an integer and x raised to any odd integer is greater than zero means that x is a positive integer, thus its lowest possible value is 1.

Hope it's clear.
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Re: x is an integer and x raised to any odd integer is greater  [#permalink]

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New post 16 Jun 2013, 21:48
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ykaiim wrote:
x is an integer and x raised to any odd integer is greater than zero; is w - z greater than 5 times the quantity 7^(x-1)-5^x?

(1) z < 25 and w=7^x
(2) x = 4



Responding to a pm:

What is the meaning of: x is an integer and x raised to any odd integer is greater than zero?
It is just a convoluted way of saying 'x is a positive integer'. Since x raised to any odd power is positive, x must be positive. x can take values 1, 2, 3, 4 and so on.

Question: Is \(w - z > 5*7^{x-1}-5^{x+1}\)

Using statement 1: Is \(7^x - 25 > 5*7^{x-1} - 5^{x+1}\)

Even though we are given that z is less than 25, let's assume it to be 25. In case we can prove that left hand side is greater than right hand side for z = 25, we can prove that left hand side will remain greater than right hand side for any other value of z. If z has a smaller value, say 9, left had side will become even larger (since a smaller number will be subtracted) than the right hand side.

Now note that \(7^x\) will be greater than \(5*7^{x - 1}\) since \(7^x\) can be written as \(7*7^{x - 1}\) (x is a positive integer).
Also note that minimum value of x is 1 so \(5^{x + 1}\) will always be equal to or larger than 25 (x is a positive integer).

So comparing the left and right hand side expressions, \(7^x\) is greater than \(5*7^{x - 1}\) and 25 is less than or equal to \(5^{x+1}\). So left hand side must be greater than the right hand side in all cases when x is a positive integer.

Answer (A)
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Re: x is an integer and x raised to any odd integer is greater  [#permalink]

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New post 05 May 2014, 04:24
paraphrasing the questions gives us, is w-z > 5.7^(x-1) - 5^(x+1). Further, x is a positive integer as an odd power of x is greater than 0.

taking statement 1, you have the question as 7^x-z > 5.7^(x-1) - 5^(x+1)

Now, for all cases where x is a positive integer, 7^x would be greater than 5.7^(x-1). Next, 5^(x+1) would always be >= 25 as the least value of x can be 1.

So, 7^x > 5.7^x-1
and z < 5^(x+1). Reversing this inequality and adding both the inequalities, we get:

7^x - z > 5.7^x-1 - 5^(x+1),

Hence answer A.
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Re: x is an integer and x raised to any odd integer is greater  [#permalink]

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New post 18 Jul 2015, 03:24
x raised to any odd integer is greater than 0 --> x = integer >0 --> x>=1

(1) --> w-z > 7^x - 5^2 = 7.7^(x-1) - 5^2 > 5.7^(x-1) - 5^2

in order to w-z> 5.7^(x-1) - 5^(x+1), x+1 >=2 --> x >=1 --> sufficient.

(2) insufficient

=> A.
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Re: x is an integer and x raised to any odd integer is greater  [#permalink]

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New post 21 Aug 2018, 09:22
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ykaiim wrote:
x is an integer and x raised to any odd integer is greater than zero; is w - z greater than 5 times the quantity 7^(x-1)-5^x?

(1) z < 25 and w = 7^x
(2) x = 4


Given: x is an integer and x raised to any odd integer is greater than zero
KEY CONCEPT: A number raised to an ODD power retains its sign.
That is, POSITIVE^ODD = POSITIVE and NEGATIVE^ODD = NEGATIVE
The given info tells us that x^ ODD = POSITIVE
So, it must be the case that x is a POSITIVE integer

Target question: Is w - z > 5[7^(x-1) - 5^x?]
Let's tidy the target question by expanding the right side of the inequality to get....
REPHRASED target question: Is w - z > 5[7^(x-1)] - 5^(x+1)?

Statement 1: z < 25 and w = 7^x
Take the REPHRASED target question and replace w with 7^x to get:
Is 7^x - z > 5[7^(x-1)] - 5^(x+1)?

IMPORTANT: Notice that, on the left side of the inequality, we're subtracting z
We're told that z < 25
So, we can help MINIMIZE the value of the left side by making z as big as possible
So, let's make z = 25 (aside: we could make z equal something really close to 25, like 24.99999999999, but let's make things easy on ourselves and make z EQUAL 25.
We get: Is 7^x - 25 > 5[7^(x-1)] - 5^(x+1)?

Now let's get like terms on the same side of the inequality.
First subtract 5[7^(x-1)] from both sides to get: Is 7^x - 5[7^(x-1)] - 25 > -5^(x+1)?
Then add 25 to both sides to get: Is 7^x - 5[7^(x-1)] > 25 - 5^(x+1)?
Factor 7^(x-1) from left side to get: Is 7^(x-1)[7 - 5] > 25 - 5^(x+1)?
Simplify left side to get: Is 7^(x-1)[2] > 25 - 5^(x+1)?

Now recognize that 7^(x-1) will be POSITIVE for all values of x.
So, 7^(x-1)[2] must be POSITIVE

Now recognize that since x is a POSITIVE integer, 5^(x+1) must be greater than or equal to 25 (since x = 1, is the smallest possible value of x)
So, 25 - 5^(x+1) must be less than or equal to zero.

So, our question becomes Is some POSITIVE number > some number that's less than or equal to zero?
The answer to this REPHRASED target question is a definitive YES
Since we can answer the REPHRASED target question with certainty, statement 1 is SUFFICIENT

Statement 2: x = 4
Since we have no information about z or w, we cannot answer the REPHRASED target question with certainty.
So, statement 2 is NOT SUFFICIENT

Answer: A

Cheers,
Brent
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Re: x is an integer and x raised to any odd integer is greater &nbs [#permalink] 21 Aug 2018, 09:22
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