x@y = 2x^(1/2) + y^2. How many unique sums of x and y result, if x@y

Answer should be C.

Here is how I did it, literally manually!
Again I am ignoring the fact that x could be fractions. Then there would be more possibilities.
The possible values of y are 0,1,2,3
For y = 0, x = 0,1,4,9,16,25,36,49 n = 8
For y = 1, x = 1, 4, 9, 16, 25, 16 , n = 6. you leave out x = 0, since 0+1 is counted already.
For y = 2, x = 1, 4, 9, 16, 25, 36, n = 6. ignore x = 0
For y = 3, x = 4, n=1 . ignore x = 0 & 1

# of Unique sums = 21

In these type of questions where we have to literally count the values try to first find the values of the constant which has max. restrictions.
Here clearly x cannot be negative because complex numbers are not considered on the gmat. Thus 2x^(1/2),x will take only positive values also y^2 will always be positive hence the complete expression will always result in non negative values.
Values for x@y = 0,1,2,3,4,5,6,7,8,9,10,11,12,13,14
Values of x = 0,1,4,9,16,25,36,49 for y =0
Values of x = 1,4,9,16,25,16 for y=1
Values of x = 1,4,9,16,25 for y=2
Values of x = 4 for y=3
total values 21
Answer C

y has to be positive but why does it have to be an integer? Why can't we have x = 0 and \(y = \sqrt{14}\)?

In that case \(x@y = 2\sqrt{x} + y^2 = 14\) (which is an integer less than 15)
In this case, x+y = \(\sqrt{14}\)

A re-think of the strategy may be required here.

I am getting 29

this is what i did...

2\sqrt{x} + y^2 < 15

x = 0
y = 0,1,2,3,-1,-2,-3
unique integer sums = -1,-2,-3,0,1,2,3

x = 1
y = 0,1,2,3,-1,-2,-3
unique integer sums = 4

x = 4
y = 0,1,2,3,-1,-2,-3
unique integer sums = 5,6,7

x = 9
y = 0,1,2,-1,-2
unique integer sums = 8,9,10,11

x = 16
y = 0,1,2,-1,-2
unique integer sums = 14,15,16,17,18

x = 25
y = 0,1,2,-1,-2
unique integer sums = 23,24,25,26,27

x = 36
y = 0,1,-1
unique integer sums = 35,36,37

x = 49
y = 0
unique integer sums = 49

total unique integer sums = 29

can someone please point out the mistake....

True. I went through the numbers.

Let's start with x=0.25
That way 2\(\sqrt{x}\) will be 1
Therefore, y can be anything from 0 to 13
x+y will be 0.25 to 13.25--->13 nos

Similarly x can be 2.25, 6.25, 12.25, 20.25, 30.25, 42.25 giving us 2, 4, 6, 4,2 different nos.

Adding all these gives us 31. The biggest option is E. 30.

And we havent even considered x as an integer yet! So if we add the options wherein x is an integer, we would have an answer way more than the options given.

I feel there is something wrong with the question.

1. Since most of us had to get down to arm-twisting tactics to solve this, we had to spend considerably more than 2 mins to solve this (Unless there is a method to this madness which we haven't been able to consider)
2. There seems to be something wrong with the options given, since the approach that you showed us fits well into the scheme of the question.

Hey Bunuel and VeritasPrepKarishma. Could you look into my concern and tell me where I am going wrong?

x@y must be less than 15 . And
2sqrX + y^2 is more than 0, so
x@y is an integer more or equal to 0 and less than 15 .
There are 15 solutions for x@y : 0, 1 , 2 ... 14

Each solution can have two combination of x and y. For each combination the sum of x + y is unique.

Therefore, 15*2= 30

There are 30 unique sums of x and y . Answer E.

it took A minimum number of 5 minutes...

Bunuel

Since there are no restrictions for x and y, there are much more than 30 possible sums for x & y, even when x@y equal an integer less than 15.
e.g. x=1/4 and y=sqrt(13), and the sum of x@y is 14, and the sum of x + y = 1/4 + sqrt(13). This can be done for many more combinations of x & y.

\(2\sqrt{x}\) = {2, 4, 6, 8, 10, 12, 14} that satisfies the inequality (<15).
----- Accordingly, x = {1, 4, 9, 16, 25, 36, 49}
\(y^2\) = {0, 1, 4, 9} that satisfies the inequality (<15).
----- Accordingly, y = {0, -1, +1, -2, +2, -3, +3}

For \(2\sqrt{x}\) = 14, \(y^2\) = 0 ; (x,y) = (49, 0) ; Sum = 49 (1 unique sum)
For \(2\sqrt{x}\) = 12, \(y^2\) = 0, 1 ; (x,y) = (36, -1), (36, 0), (36, 1) ; Sums: 35, 36, 37 (3 unique sums)
For \(2\sqrt{x}\) = 10, \(y^2\) = 0, 1, 4 ; (x,y) = (25; -2, -1, 0, 1, 2) ; Sums: 23~27 (5 unique sums)
For \(2\sqrt{x}\) = 8, \(y^2\) = 0, 1, 4 ; (x,y) = (16; -2, -1, 0, 1, 2) ; Sums: 14~18 (5 unique sums)
For \(2\sqrt{x}\) = 6, \(y^2\) = 0, 1, 4 ; (x,y) = (9; -2, -1, 0, 1, 2) ; Sums: 7~11 (5 unique sums)
For \(2\sqrt{x}\) = 4, \(y^2\) = 0, 1, 4, 9 ; (x,y) = (4; -3, -2, -1, 0, 1, 2, 3) ; Sums: 1~7 (7 unique sums)
For \(2\sqrt{x}\) = 2, \(y^2\) = 0, 1, 4, 9 ; (x,y) = (1; -3, -2, -1, 0, 1, 2, 3) ; Sums: -3~4 (7-3=4 unique sums; 1, 2, 3 overlap with the sums right above.)

Therefore, there are 1+3+5+5+5+7+4=30 unique sums. E.

I dont see how we can count the unique sums if x and y are not restricted to integers.
Here, we can not assume that x and y are integers.
x@y can be integer while x and y are not integers.
example x=9/64 and y=1/2 leads to x@y=1

Experts Bunuel VeritasPrepKarishma, can you please advise?

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