If x and y are positive integers and 615 + x^2 = 2^y, what is the valu

Question

If x and y are positive integers and  615 + x^2 = 2^y , what is the value of x?

A. 53 
B. 55
C. 57
D. 59
E. 61

Are You Up For the Challenge: 700 Level Questions

Kinshook · Answer

Asked: If x and y are positive integers and  615 + x^2 = 2^y , what is the value of x?

x^2 = 2^y - 615 
 2^y > 615

A perfect square has unit digits = {0,1,4,9,6,5}

2^{10} = 1024; 1024 - 615 = 409 ; Not a perfect square
 2^{11} = 2048; 2048 - 615 = 1433 ; Not a perfect square
 2^{11} = 4096; 4096 - 615 = 3481 = 59^2 ; A perfect square

x = 59

IMO D

Kinshook · Answer

Asked: If x and y are positive integers and  615 + x^2 = 2^y , what is the value of x?

A. 53 
615 + 53^2 = 615 + 2809 = 3424; Not of form 2^y
B. 55
615 + 55^2 = 615 + 3025 = 3640; Not of form 2^y
C. 57
615 + 57^2 = 615 + 3249 = 3864; Not of form 2^y
D. 59
615 + 59^2 = 615 + 3481 = 4096; Of the form 2^y; Correct Answer
E. 61
615 + 61^2 = 615 + 3721 = 4336; Not of form 2^y

IMO D

yashikaaggarwal · Answer

615 + x^2 = 2^y
X has to be an integer whose square root when added with 615 will give out only 2 as factors.
Power of 2 whose product gives a no. Bigger than 615 = 2^10 = 1024 (2^9 = 512, is less than 615)
=> 2^10-615 => 1024-615 = 409 = X^2 => X is not an integer.
Similarly,
=> 2^11-615 => 2048-615 = 1433 = X^2 => X is not an integer.
=> 2^12-615 => 4096-615 = 3481 = X^2 => X = 59 (an integer)

Answer is D

Posted from my mobile device

GMATWhizTeam · Answer

Solution

• We have  615 + x^2 = 2^y 
• Now, if we look at the answer options, we have possible values of x as
 o  50 < x < 61   
  ⟹ 2500 < x^2 < 3721 
  ⟹ 2500 + 615 < x^2 + 615 < 3721 + 615 
  ⟹ 3115 < 2^y < 4336  
• So, only possible value of  2^y  , given that y is a positive integer, in the above range = 4096
• Thus,  x^2 = 4096 – 615 = 3481 = 59^2 ⟹ x = 59  
Thus, the correct answer is  Option D.

Lipun · Answer

The given options are in the range of 50^2 + 615 to 60^2 + 615. 
The estimated range of 2^y will be 3100 to 4200. The only value satisfying 2^y in that range is 4096.

So, the answer must be closer to 60. Since 60^2 = 3600 and 3600+615 > 4096, so we start checking with 59

59^2 = 3481. Adding 615 to it we obtain 4096

Ans: 59

ScottTargetTestPrep · Answer

Solution:
 
First, let’s approximate the value on the left hand side of the equation by using the choices given. If x = 53, 615 + x^2 is approximately 600 + 2500 = 3100. If x = 61, 615 + x^2 is approximately 600 + 3600 = 4200. Since the only power of 2 that is between 3100 and 4200 is 2^12 = 4096, we see that y must be 12. So we can solve for x as follows:

615 + x^2 = 2^12

615 + x^2 = 4096

x^2 = 3481

x = √3481 = 59

Answer: D

bumpbot · Answer

Automated notice from GMAT Club BumpBot: 

A member just gave Kudos to this thread, showing it’s still useful. I’ve bumped it to the top so more people can benefit. Feel free to add your own questions or solutions.

 This post was generated automatically.

If x and y are positive integers and 615 + x^2 = 2^y, what is the valu

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions

If x and y are positive integers and 615 + x^2 = 2^y, what is the valu

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards