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PKN
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Given p=|q| and q=3^a-(27^40+27^8), which of the following values of ' 14 Jul 2018, 07:32
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

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45% (medium)

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70% (02:22) correct 30% (02:30) wrong based on 966 sessions

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Given p=|q| and \(q=3^a-(27^{40}+27^8)\), which of the following values of 'a' yields the least value for p ?

(A) 120
(B) 140
(C) 180
(D) 185
(E) 190
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A
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LeenaPG
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Given p=|q| and q=3^a-(27^40+27^8), which of the following values of ' 27 Jul 2018, 06:01
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Hi,

I think it could also be solved in the following way:
q=3^a−(27^40+27^8)=3^a- ((3^3)^40 + (3^3)^8

= 3^a - (3^120 + 3^24)

= 3^a - 3^120 - 3^24

if a = 120 then the result would be - 3^24. For all other values, it will be greater than this.

So, Answer is A.
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Given p=|q| and q=3^a-(27^40+27^8), which of the following values of ' 14 Jul 2018, 07:48
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PKN
Given p=|q| and \(q=3^a-(27^{40}+27^8)\), which of the following values of 'a' yields the least value for p ?

(A) 120
(B) 140
(C) 180
(D) 185
(E) 190
Show more

since p is always positive and p = |q|, we have to find the least |q|
now \(q=3^a-(27^{40}+27^8)=3^a-27^8(27^{32}+1)=3^a-27^8(27^{32}=3^a-27^{40}=3^a-3^{120}\),
so a ~ 120

A
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Given p=|q| and q=3^a-(27^40+27^8), which of the following values of ' 23 Jul 2018, 17:37
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chetan2u
PKN
Given p=|q| and \(q=3^a-(27^{40}+27^8)\), which of the following values of 'a' yields the least value for p ?

(A) 120
(B) 140
(C) 180
(D) 185
(E) 190

since p is always positive and p = |q|, we have to find the least |q|
now \(q=3^a-(27^{40}+27^8)=3^a-27^8(27^{32}+1)=3^a-27^8(27^{32}=3^a-27^{40}=3^a-3^{120}\),
so a ~ 120

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what happened to the +1 after the 27^{32}?
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Given p=|q| and q=3^a-(27^40+27^8), which of the following values of ' 23 Jul 2018, 18:34
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nati1516
chetan2u
PKN
Given p=|q| and \(q=3^a-(27^{40}+27^8)\), which of the following values of 'a' yields the least value for p ?

(A) 120
(B) 140
(C) 180
(D) 185
(E) 190

since p is always positive and p = |q|, we have to find the least |q|
now \(q=3^a-(27^{40}+27^8)=3^a-27^8(27^{32}+1)=3^a-27^8(27^{32}=3^a-27^{40}=3^a-3^{120}\),
so a ~ 120

A

what happened to the +1 after the 27^{32}?
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Hi....
Since 1 is very very small as compared to 27^{32}, we can discard 1 as we are looking at an approximate value
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Given p=|q| and q=3^a-(27^40+27^8), which of the following values of ' 28 Jul 2018, 01:27
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Bunuel pushpitkc Bunuel niks18 gmatbusters KarishmaB

I have a small query regarding the penultimate step by chetan2u as below:
Quote:
\(3^a-3^{120}\),
so a ~ 120
Show more

Here is the question stem again which asks for a definite value and not approximation.

Quote:
Given p=|q| and \(q=3^a-(27^{40}+27^8)\), which of the following values of 'a' yields the least value for p ?
Show more

I believe chetan2u took the highlighted text to infer that \(3^a\) must be a positive value.

What I did not get was: if a is actually equal to 120 then the final value of q will be 0 and p will be 0 , which voilates Q stem.
Can you suggest why we did not take next positive value i.e. 140 as correct OA?
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Given p=|q| and q=3^a-(27^40+27^8), which of the following values of ' 28 Jul 2018, 02:56
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Hi

Query1:

I believe chetan2u took the highlighted text to infer that \(3^a\) must be a positive value.. it is not to be inferred from the question stem.

3^a is always positive- 3^a can never be negative and 3^a tends to zero only if a approaches negative infinity.

Concept: (positive)^x, is always positive, it can not be negative. (positive)^x tends to zero only if a approaches negative infinity.



Query2: if a is actually equal to 120 then the final value of q will be 0 and p will be 0 , which violates Q stem
I am afraid, I couldn't understand this point. How it violates the Question stem.- please explain the query further to let me expand.

In fact , p=|q|. Hence the minimum value of p can be equal to zero.( it cannot be negative). So , as explained in above posts, p approaches zero, when a is 120. (out of the given 5 options).


adkikani

gmatbusters

I have a small query regarding the penultimate step by chetan2u as below:
Quote:
\(3^a-3^{120}\),
so a ~ 120

Here is the question stem again which asks for a definite value and not approximation.

Quote:
Given p=|q| and \(q=3^a-(27^{40}+27^8)\), which of the following values of 'a' yields the least value for p ?

I believe chetan2u took the highlighted text to infer that \(3^a\) must be a positive value.

What I did not get was: if a is actually equal to 120 then the final value of q will be 0 and p will be 0 , which voilates Q stem.
Can you suggest why we did not take next positive value i.e. 140 as correct OA?
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Given p=|q| and q=3^a-(27^40+27^8), which of the following values of ' 28 Jul 2018, 03:00
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adkikani
Bunuel pushpitkc Bunuel niks18 gmatbusters KarishmaB

I have a small query regarding the penultimate step by chetan2u as below:
Quote:
\(3^a-3^{120}\),
so a ~ 120

Here is the question stem again which asks for a definite value and not approximation.

Quote:
Given p=|q| and \(q=3^a-(27^{40}+27^8)\), which of the following values of 'a' yields the least value for p ?

I believe chetan2u took the highlighted text to infer that \(3^a\) must be a positive value.

What I did not get was: if a is actually equal to 120 then the final value of q will be 0 and p will be 0 , which violates Q stem.
Can you suggest why we did not take next positive value i.e. 140 as correct OA?
Show more

Hi adkikani,

As you know, |q|=q when q>0
|q|=-q when q<0
So, p=|q|=positive
Hence, determining the least value of 'p' is same as determining the least value(minimum) of q.

Given \(q=3^a-(27^{40}+27^8)=3^a-(3^{120}-3^{24})=3^a-3^{120}-3^{24}\)--------------(1)

adkikani

What I did not get was: if a is actually equal to 120 then the final value of q will be 0 and p will be 0 , which violates Q stem.
Show more

when a=120, \(q=3^{120}-3^{120}+3^{24}\)=\(-{3^{24}}\neq0\)

Quote:
\(3^a-3^{120}\),
so a ~ 120
Here is the question stem again which asks for a definite value and not approximation.
Show more

Here a=120 is not approximated. When a=120, the given expression has the least value of q, i.e. \(-3^{24}\). So, \(p=|q|=-q=-(-3^{24})=3^{24}\)

Quote:
Can you suggest why we did not take next positive value i.e. 140 as correct OA?
Show more

Let's take a=140, so (1) becomes, \(q=3^{140}-3^{120}-3^{24}\). As \(3^{140}-3^{120}\)is positive, so \(|q|> 3^{24}\)
But we want the least value. Therefore, we have to discard a=140.

Quote:
I believe chetan2u took the highlighted text to infer that \(3^a\) must be a positive value.Given p=|q| and \(q=3^a-(27^{40}+27^8)\), which of the following values of 'a' yields the least value for p ?
Show more

No, the highlighted portion signifies 'p' must be positive. (q<0 (say -5) implies p=|q|=-q=-(-5)=positive; q>0 implies p=|q|=positive)

P.S:- We can have value of the expression(|q|) less than \(3^{24}\) but we are told to determine the least value of 'p' from the given 5 answer options.(a=120, 140,180,185, and 190)
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Given p=|q| and q=3^a-(27^40+27^8), which of the following values of ' 28 Jul 2018, 03:45
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adkikani
Bunuel pushpitkc Bunuel niks18 gmatbusters KarishmaB

I have a small query regarding the penultimate step by chetan2u as below:
Quote:
\(3^a-3^{120}\),
so a ~ 120

Here is the question stem again which asks for a definite value and not approximation.

Quote:
Given p=|q| and \(q=3^a-(27^{40}+27^8)\), which of the following values of 'a' yields the least value for p ?

I believe chetan2u took the highlighted text to infer that \(3^a\) must be a positive value.

What I did not get was: if a is actually equal to 120 then the final value of q will be 0 and p will be 0 , which voilates Q stem.
Can you suggest why we did not take next positive value i.e. 140 as correct OA?
Show more

Most of the points have been made amply clear in above posts. However, the point on approximation..
What is the least value of p, it is 0 as p=|q|
And at a=120, it is not 0, it is 3^120-(3^120+3^24)=-3^24
So p =|-3^24|=3^24

So that is not equal to 0..
But are we looking for 'a' to get the least value of p...NO then a will be 120.xyz, something in decimals.
We are looking for the least value for the FOLLOWING values
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Given p=|q| and q=3^a-(27^40+27^8), which of the following values of ' 28 Sep 2025, 07:12
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But if we plug in 3 ^ 140, 3^140 - 3^120 would give 3^20 (positive)

therefore we would be remained with 3^20- 3^24, so would'nt this option be smaller?
LeenaPG
Hi,

I think it could also be solved in the following way:
q=3^a−(27^40+27^8)=3^a- ((3^3)^40 + (3^3)^8

= 3^a - (3^120 + 3^24)

= 3^a - 3^120 - 3^24

if a = 120 then the result would be - 3^24. For all other values, it will be greater than this.

So, Answer is A.
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Given p=|q| and q=3^a-(27^40+27^8), which of the following values of ' 28 Sep 2025, 07:22
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ok, I made a dumb mistake, in exponents there is no rule if bases are same and are getting subtracted, so rather it would be 3^{140} - 3^120 = 3^{120}(3^{20} - 1).
iurequi
But if we plug in 3 ^ 140, 3^140 - 3^120 would give 3^20 (positive)

therefore we would be remained with 3^20- 3^24, so would'nt this option be smaller?

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