GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 20 Oct 2019, 12:28 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

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.  M26-21

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics
Author Message
TAGS:

Hide Tags

Math Expert V
Joined: 02 Sep 2009
Posts: 58434

Show Tags

1
3 00:00

Difficulty:   25% (medium)

Question Stats: 69% (01:01) correct 31% (01:13) wrong based on 203 sessions

HideShow timer Statistics

What is the smallest positive integer $$k$$ such that $$126*\sqrt{k}$$ is the square of a positive integer?

A. 14
B. 36
C. 144
D. 196
E. 441

_________________
Math Expert V
Joined: 02 Sep 2009
Posts: 58434

Show Tags

Official Solution:

What is the smallest positive integer $$k$$ such that $$126*\sqrt{k}$$ is the square of a positive integer?

A. 14
B. 36
C. 144
D. 196
E. 441

$$126=2*3^2*7$$, so in order $$126*\sqrt{k}$$ to be a square of an integer, $$\sqrt{k}$$ must complete the powers of 2 and 7 to even number, so the least value of $$\sqrt{k}$$ must equal $$2*7=14$$, which makes the least value of $$k$$ equal to $$14^2=196$$.

_________________
Director  G
Joined: 02 Sep 2016
Posts: 649

Show Tags

Bunuel wrote:
Official Solution:

What is the smallest positive integer $$k$$ such that $$126*\sqrt{k}$$ is the square of a positive integer?

A. 14
B. 36
C. 144
D. 196
E. 441

$$126=2*3^2*7$$, so in order $$126*\sqrt{k}$$ to be a square of an integer, $$\sqrt{k}$$ must complete the powers of 2 and 7 to even number, so the least value of $$\sqrt{k}$$ must equal $$2*7=14$$, which makes the least value of $$k$$ equal to $$14^2=196$$.

Hi Bunuel

Your help here would be great.

Please guide me where I went wrong.

126*\sqrt{k}= (a)^2

Here a is the number we are talking about in the question stem.

126= 2*3^2*7

So 2*3^2*7*\sqrt{k}= a^2

Squaring both the sides

2^2*3^4*7^2*\sqrt{k}= a

So the lowest we can plug in is 36 (6^2) as we already have all the numbers with even powers.
Math Expert V
Joined: 02 Sep 2009
Posts: 58434

Show Tags

1
Shiv2016 wrote:
Bunuel wrote:
Official Solution:

What is the smallest positive integer $$k$$ such that $$126*\sqrt{k}$$ is the square of a positive integer?

A. 14
B. 36
C. 144
D. 196
E. 441

$$126=2*3^2*7$$, so in order $$126*\sqrt{k}$$ to be a square of an integer, $$\sqrt{k}$$ must complete the powers of 2 and 7 to even number, so the least value of $$\sqrt{k}$$ must equal $$2*7=14$$, which makes the least value of $$k$$ equal to $$14^2=196$$.

Hi Bunuel

Your help here would be great.

Please guide me where I went wrong.

126*\sqrt{k}= (a)^2

Here a is the number we are talking about in the question stem.

126= 2*3^2*7

So 2*3^2*7*\sqrt{k}= a^2

Squaring both the sides

2^2*3^4*7^2*\sqrt{k}= a

So the lowest we can plug in is 36 (6^2) as we already have all the numbers with even powers.

If you square $$2*3^2*7*\sqrt{k}= a^2$$ you get $$2^2*3^4*7^2*k= a^4$$.
_________________
Director  G
Joined: 02 Sep 2016
Posts: 649

Show Tags

1
Bunuel wrote:
Shiv2016 wrote:
Bunuel wrote:
Official Solution:

What is the smallest positive integer $$k$$ such that $$126*\sqrt{k}$$ is the square of a positive integer?

A. 14
B. 36
C. 144
D. 196
E. 441

$$126=2*3^2*7$$, so in order $$126*\sqrt{k}$$ to be a square of an integer, $$\sqrt{k}$$ must complete the powers of 2 and 7 to even number, so the least value of $$\sqrt{k}$$ must equal $$2*7=14$$, which makes the least value of $$k$$ equal to $$14^2=196$$.

Hi Bunuel

Your help here would be great.

Please guide me where I went wrong.

126*\sqrt{k}= (a)^2

Here a is the number we are talking about in the question stem.

126= 2*3^2*7

So 2*3^2*7*\sqrt{k}= a^2

Squaring both the sides

2^2*3^4*7^2*\sqrt{k}= a

So the lowest we can plug in is 36 (6^2) as we already have all the numbers with even powers.

If you square $$2*3^2*7*\sqrt{k}= a^2$$ you get $$2^2*3^4*7^2*k= a^4$$.

Then k= a^4/(2^2*3^4*7^2)

As we do not know the value of a, we will have to take a as 2^2*7^2 because the power of a is 4. 2,3,7, and k are its prime factors and hence will take the power of 4.

These silly mistakes are going out of my hand.
Manager  S
Joined: 25 Jul 2017
Posts: 93

Show Tags

Bunuel wrote:
What is the smallest positive integer $$k$$ such that $$126*\sqrt{k}$$ is the square of a positive integer?

A. 14
B. 36
C. 144
D. 196
E. 441

This question can be easily solved by factorisation.

126*√k
=> 2*3*3*7*√k
=> 3^2*14*√k

i.e. we need another 14 to make it a square of positive number
i.e. √k=14
K= 14^2 = 196
D is the answer. Re: M26-21   [#permalink] 15 Aug 2018, 20:56
Display posts from previous: Sort by

M26-21

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics

Moderators: chetan2u, Bunuel

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne  