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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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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Bunuel pushpitkc Bunuel niks18 gmatbusters KarishmaB

I have a small query regarding the penultimate step by chetan2u as below:


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



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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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)



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



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



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.



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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