GMAT Question of the Day: Daily via email | Daily via Instagram New to GMAT Club? Watch this Video

 It is currently 26 Jan 2020, 03:58

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

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

Author Message
TAGS:

### Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 60646
If (5/4)^(-n) < 16^(-1). What is the least integer value of n?  [#permalink]

### Show Tags

01 Aug 2017, 01:06
7
28
00:00

Difficulty:

95% (hard)

Question Stats:

30% (02:40) correct 70% (02:38) wrong based on 242 sessions

### HideShow timer Statistics

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

_________________
Math Expert
Joined: 02 Aug 2009
Posts: 8336
If (5/4)^(-n) < 16^(-1). What is the least integer value of n?  [#permalink]

### Show Tags

01 Aug 2017, 06:07
4
8
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
_________________
##### General Discussion
Retired Moderator
Joined: 25 Feb 2013
Posts: 1156
Location: India
GPA: 3.82
Re: If (5/4)^(-n) < 16^(-1). What is the least integer value of n?  [#permalink]

### Show Tags

25 Sep 2017, 11:02
1
1
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?
Manager
Joined: 24 Jun 2017
Posts: 115
Re: If (5/4)^(-n) < 16^(-1). What is the least integer value of n?  [#permalink]

### Show Tags

25 Sep 2017, 11:11
1
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
Director
Joined: 13 Mar 2017
Posts: 727
Location: India
Concentration: General Management, Entrepreneurship
GPA: 3.8
WE: Engineering (Energy and Utilities)
If (5/4)^(-n) < 16^(-1). What is the least integer value of n?  [#permalink]

### Show Tags

03 Sep 2017, 11:39
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..
Intern
Joined: 16 Apr 2017
Posts: 17
Re: If (5/4)^(-n) < 16^(-1). What is the least integer value of n?  [#permalink]

### Show Tags

25 Sep 2017, 09:54
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
Manager
Joined: 24 Jun 2017
Posts: 115
Re: If (5/4)^(-n) < 16^(-1). What is the least integer value of n?  [#permalink]

### Show Tags

25 Sep 2017, 14:05
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
Senior Manager
Joined: 04 Apr 2015
Posts: 254
GMAT 1: 650 Q49 V31
GPA: 3.59
Re: If (5/4)^(-n) < 16^(-1). What is the least integer value of n?  [#permalink]

### Show Tags

03 Oct 2017, 21:07
mikemcgarry can you please explain this Magoosh question?
Magoosh GMAT Instructor
Joined: 28 Dec 2011
Posts: 4476
Re: If (5/4)^(-n) < 16^(-1). What is the least integer value of n?  [#permalink]

### Show Tags

05 Oct 2017, 09:20
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
_________________
Mike McGarry
Magoosh Test Prep

Education is not the filling of a pail, but the lighting of a fire. — William Butler Yeats (1865 – 1939)
Senior Manager
Joined: 04 Apr 2015
Posts: 254
GMAT 1: 650 Q49 V31
GPA: 3.59
Re: If (5/4)^(-n) < 16^(-1). What is the least integer value of n?  [#permalink]

### Show Tags

05 Oct 2017, 09:27
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

Senior Manager
Joined: 02 Apr 2014
Posts: 460
Location: India
Schools: XLRI"20
GMAT 1: 700 Q50 V34
GPA: 3.5
If (5/4)^(-n) < 16^(-1). What is the least integer value of n?  [#permalink]

### Show Tags

07 Jan 2018, 19:20
$$(4/5)^n < 1/16$$

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
Intern
Joined: 07 Jun 2018
Posts: 19
Location: India
GPA: 3.9
WE: Marketing (Insurance)
If (5/4)^(-n) < 16^(-1). What is the least integer value of n?  [#permalink]

### Show Tags

14 Sep 2018, 09:04
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. as the question was definitely not an easy one to crack.
Non-Human User
Joined: 09 Sep 2013
Posts: 14015
Re: If (5/4)^(-n) < 16^(-1). What is the least integer value of n?  [#permalink]

### Show Tags

29 Nov 2019, 12:05
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
Re: If (5/4)^(-n) < 16^(-1). What is the least integer value of n?   [#permalink] 29 Nov 2019, 12:05
Display posts from previous: Sort by