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

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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

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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

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

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