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# M26-21

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Math Expert
Joined: 02 Sep 2009
Posts: 58434

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16 Sep 2014, 01:25
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25% (medium)

Question Stats:

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

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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
Joined: 02 Sep 2009
Posts: 58434

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16 Sep 2014, 01:25
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 $$14^2=196$$.

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Director
Joined: 02 Sep 2016
Posts: 649

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06 Jan 2017, 04:56
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
Joined: 02 Sep 2009
Posts: 58434

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06 Jan 2017, 05:29
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
Joined: 02 Sep 2016
Posts: 649

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06 Jan 2017, 22:27
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
Joined: 25 Jul 2017
Posts: 93

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15 Aug 2018, 20:56
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
Re: M26-21   [#permalink] 15 Aug 2018, 20:56
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# M26-21

Moderators: chetan2u, Bunuel