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If (5/4)^(-n) < 16^(-1). What is the least integer value of n?

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If (5/4)^(-n) < 16^(-1). What is the least integer value of n?  [#permalink]

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New post 01 Aug 2017, 00:06
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If (5/4)^(-n) < 16^(-1). What is the least integer value of n?  [#permalink]

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New post 01 Aug 2017, 05:07
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Bunuel wrote:
If \((\frac{5}{4})^{-n} < 16^{-1}\). What is the least integer value of n?

A. 12
B. 13
C. 14
D. 15
E. 16


if you counter such Q, an approximation way would be..

\((\frac{5}{4})^{-n} < 16^{-1}..........(\frac{4*2}{5*2})^n<\frac{1}{16}......2^{3n}*16<10^n....2^{3n+4}<10^n\)

now\(2^{10}=1024\) and \(10^3 = 1000\) ..nearly equal so lets convert ..

If you could get some even higher close values, the answer will be even closer
\(1024^{\frac{3n+4}{10}}<1000^{\frac{n}{3}}\)...

so \(\frac{3n+4}{10}<\frac{n}{3}........9n+12<10n....n>12\)
next integer value after 12 is 13....
so ans 13
B
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If (5/4)^(-n) < 16^(-1). What is the least integer value of n?  [#permalink]

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New post 03 Sep 2017, 10:39
chetan2u wrote:
Bunuel wrote:
If \((\frac{5}{4})^{-n} < 16^{-1}\). What is the least integer value of n?

A. 12
B. 13
C. 14
D. 15
E. 16


if you counter such Q, an approximation way would be..

\((\frac{5}{4})^{-n} < 16^{-1}..........(\frac{4*2}{5*2})^n<\frac{1}{16}......2^{3n}*16<10^n....2^{3n+4}<10^n\)

now\(2^{10}=1024\) and \(10^3 = 1000\) ..nearly equal so lets convert ..

If you could get some even higher close values, the answer will be even closer
\(1024^{\frac{3n+4}{10}}<1000^{\frac{n}{3}}\)...

so \(\frac{3n+4}{10}<\frac{n}{3}........9n+12<10n....n>12\)
next integer value after 12 is 13....
so ans 13
B


The options are so close, How can one say that on taking 1024 ~ 1000 , one will get correct answer..
one might get n>13 or n>11 instead of n>12 in some other question of this type....

We need to find some other approach..
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Re: If (5/4)^(-n) < 16^(-1). What is the least integer value of n?  [#permalink]

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New post 25 Sep 2017, 08:54
I would say taking logs on both sides would be better.

On taking logs, we get

(n+2)*log4 < n*log 5

=> (n+2)/n < log5/log4

Now log5 = 1-log2 = 1-.301 =.699
log4 = 2*log2 =0.602

=> (n+2)/n = 0.699/0.602

Putting n=12, we know that the inequality is not satisfied. Because LHS = 7/6 whereas RHS is less than 7/6 (0.699/0.602 is less than 0.7/0.6). Hence next integral value of n is correct
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Re: If (5/4)^(-n) < 16^(-1). What is the least integer value of n?  [#permalink]

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New post 25 Sep 2017, 10:02
Bunuel wrote:
If \((\frac{5}{4})^{-n} < 16^{-1}\). What is the least integer value of n?

A. 12
B. 13
C. 14
D. 15
E. 16


The question stem can be written as \((\frac{4}{5})^n < \frac{1}{16}\)

Now we know that \(16 = 2^4\) so if we can make \((\frac{4}{5})\) some approximate form of \(2\), then our job might become easier.

\((\frac{4}{5})^3 = \frac{64}{125}\) this is slightly than greater \(\frac{1}{2}\)

Hence \((\frac{4}{5})^3 = \frac{1}{2}\),

and \((\frac{1}{2})^4 = \frac{1}{16}\)

Substitute the value of \(\frac{1}{2}\) to get \((\frac{4}{5})^{12}\) but this will be greater than \(\frac{1}{16}\).

Hence to make it lower we will have to multiply \((\frac{4}{5})^{12}\) by a lower value \(\frac{4}{5}\)

Therefore \((\frac{4}{5})^{13}<\frac{1}{16}\)

So \(n=13\)

Option B

P.S: This is a very tough question and probably beyond the scope of GMAT. Unless you have solved similar questions, it will be very difficult to solve it in real GMAT within 2 minutes.

Bunuel Do you have repository of similar kind of questions?
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Re: If (5/4)^(-n) < 16^(-1). What is the least integer value of n?  [#permalink]

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New post 25 Sep 2017, 10:11
One method (gmat style :))
(5/4)^(−n)<16^(−1)
(4/5)^n<1/2^4
so left side should be less than 1/2

we raise to cube
(4/5)^3=64/125 and multiply right side by 64 just to compare, makes 64/128, so the inequality is still not satisfied but just a bit, it's clear that if multiply the left side by 4/5 one more time the inequality will be respected
But as we have 4 power on the right 1/2^4
then for the left
(4/5)^12 *4/5 which is 13
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Re: If (5/4)^(-n) < 16^(-1). What is the least integer value of n?  [#permalink]

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New post 25 Sep 2017, 13:05
1
And another approach for math geeks, solving through logs, I always recommend to use ln (natural log) just memorize 6 numbers by heart (ln for 2 3 5 6 7 and 10) you will cove most of that kind of questions https://gmatclub.com/forum/which-of-the-following-is-the-largest-163644.html#p1930561
(5/4)^(−n)<16^(−1)
ln (4/5)^(n)< ln 2^(−4)
as per log property
n *ln (4/5)< -4 * ln 2
another property
n *(ln 4 - ln5) < -4 * ln 2
n* (1.4 - 1.6) < - 4 * 0.7
n * (-0.2) < - 2.8
then
n < - 2.8/ -0.2 =>> n < 14 so 13
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Re: If (5/4)^(-n) < 16^(-1). What is the least integer value of n?  [#permalink]

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New post 03 Oct 2017, 20:07
mikemcgarry can you please explain this Magoosh question?
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Re: If (5/4)^(-n) < 16^(-1). What is the least integer value of n?  [#permalink]

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New post 05 Oct 2017, 08:20
StrugglingGmat2910 wrote:
mikemcgarry can you please explain this Magoosh question?

Dear StrugglingGmat2910,

I'm happy to respond. :-)

My friend, this was posted as a Magoosh question, but I don't believe it's one that we still have in our Product. This is a very hard question, one that is probably best solved with math that is completely beyond the GMAT. I am going to say that you don't need to worry about this question at all for the GMAT.

Does this make sense?
Mike :-)
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Re: If (5/4)^(-n) < 16^(-1). What is the least integer value of n?  [#permalink]

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New post 05 Oct 2017, 08:27
mikemcgarry wrote:
StrugglingGmat2910 wrote:
mikemcgarry can you please explain this Magoosh question?

Dear StrugglingGmat2910,

I'm happy to respond. :-)

My friend, this was posted as a Magoosh question, but I don't believe it's one that we still have in our Product. This is a very hard question, one that is probably best solved with math that is completely beyond the GMAT. I am going to say that you don't need to worry about this question at all for the GMAT.

Does this make sense?
Mike :-)

thank you sir for your reply
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If (5/4)^(-n) < 16^(-1). What is the least integer value of n?  [#permalink]

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New post 07 Jan 2018, 18:20
\((4/5)^n < 1/16\)

as both 4/5 and 1/16 are positive

we take 4th root on both sides

\((4/5)^{n/4} < 1/2\)

let \(k = n/4\)

when k = 3, \((4/5)^3\) = \(64/125\) is slightly greater than 1/2,

so least value of n for which \((4/5)^n < 1/16\) is n = 3(4) + 1 = 13
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If (5/4)^(-n) < 16^(-1). What is the least integer value of n?  [#permalink]

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New post 14 Sep 2018, 08:04
Bunuel wrote:
If \((\frac{5}{4})^{-n} < 16^{-1}\). What is the least integer value of n?

A. 12
B. 13
C. 14
D. 15
E. 16


\((\frac{5}{4})^{-n} < 16^{-1}\)
=> A = \((\frac{4}{5})^n*2^4 < 1\)

For n = 0, A =16; n = 1, A = 16*0.8.
So, as the value of n increases, A decreases and as n is increased by 1, A decreases by 80 %. We need to look for the value of n for which A >1 and A*80% less than 1.

As the power of 2 is 4, I was looking to test multiple of 4 for n while solving. So, I decided to first test with n = 12; from the list of option. As it will bit simplify A.
For n = 12, A = \((\frac{4}{5})^{12}*2^4\) = \((\frac{4^3}{5^3})^4*2^4\) = \((\frac{64}{125})^4*2^4\)
= \((\frac{128}{125})^4\)
A will be slightly greater than 1. But, definitely 80% of A will be just greater than 0.8. So, if n=13, value of A will be around 0.8 and it is the least integer value for which the inequality satisfies.
Hence, B. I might be a bit lucky. :cool: as the question was definitely not an easy one to crack.
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