If x and y are integers and neither x nor y is divisible by 3, which

Question

If x and y are integers and neither x nor y is divisible by 3, which of the following could be the value of  x^2+y^2 ? 
 
A. 333
B. 334
C. 336
D. 337
E. 338

chandnik3 · Accepted Answer

Consider the squares of the integers that are not divisible by 3:
1^2 = 1, 2^2 = 4, 4^2 = 16, 5^2 = 25, 7^2 = 49, 8^2=64,
They all have a remainder of 1 when they are divided by 3.
Thus, the sum of the squares of two integers which are not divisible by 3 must have a remainder of 2( remainder of x=1 + remainder of y=1) when it is divided by 3.

The only answer choice having this property is 338.

andrenoble · Answer

Please explain the OA. I've run the calculation in Excel and the only number close to this range is 338 (49 + 289), while it's not among the answers.
Thank you

Bunuel · Answer

You're correct, x^2 + y^2 = 335 does not yield an integer solution, whereas 338 does, with x=y=13 for instance. I've edited the options to include 338 and remove 335. Thanks!

Anne_ · Answer

If x and y are integers and neither x nor y is divisible by 3, there are 3 possible combinations for x & y
x = (3x + 1) & y = (3x + 1)
x = (3x + 1) & y = (3x + 2) - Note switching form between x & y would give us the same calculation at end.
x = (3x + 2) & y = (3x + 2)

x^2 + y^2 = (3x + 1)^2 + (3x + 1)^2 = (9x^2 + 6x + 1) + (9x^2 + 6x + 1) = (3a + 1) + (3b + 1) = 3(a+b) + 2
Observe the first 2 terms for each part, those are divisible by 3.
Similarly for other two combinations:
x^2 + y^2 = (3x + 1)^2 + (3x + 2)^2 = (9x^2 + 6x + 1) + (9x^2 + 12x + 4) = (3a + 1) + (3b + 4) = 3(a+b+1) + 2
x^2 + y^2 = (3x + 2)^2 + (3x + 2)^2 = (9x^2 + 12x + 4) + (9x^2 + 12x + 4) = (3a + 4) + (3b + 4) = 3(a+b+2) + 2

So all combinations gives us the result in 3k + 2 form. From the answer choices:
A. 333 - Is in 3k form, not possible
B. 334 - Is in 3k+1 form, not possible
C. 336 - Is in 3k form, not possible
D. 337 - Is in 3k+1 form, not possible
E. 338 - Is in 3k+2 form, possible (CORRECT CHOICE)

Kinshook · Answer

If x and y are integers and neither x nor y is divisible by 3, which of the following could be the value of  x^2+y^2 ?

x & y are of the form 3k+1 or 3k-1; where k is an integer
xˆ2 & yˆ2 are of the form 3*Integer + 1 
xˆ2 + yˆ2 is of the form 3*Integer + 2

Only 338 is of the form xˆ2 + yˆ2 = 3*Integer + 2
338 = 3*112 + 2

IMO E

Yoaerez · Answer

I am not that smart so here's a simple way that only took me 2 minutes:

x or y have to be less than 20 (20^2 = 400 and that's not in the answer choices).

x and y can't be divisible by 3 so we're left with any of these: 1,2,4,5,7,8,10,11,13,14,16,17,19 as our choices. our answer choices are all in the 333-338 range so we're gonna have to use some big numbers.

Let's run some squares of the big numbers to check whether a smaller square can accompany them:

19 is out since a quicl calculation shows it's above 340.

333-(17^2) = 44. 
so 338-(17^2) = 44+5 = 49.

49 is the square of 7, so (17^2) + (7^2) = 338. ANSWER CHOICE E.

Let's run some other numbers just for fun and to see the thought process:

333-(16^2) = 77.
338-(16^2) = 82.

only number in that range that has a square root is 81, but 9 is divisible by 3 so the answer can't be (16^2)+(9^2)

333-(15^2) = 108.
338-(15^2) = 113.

no number has a square root in that range.

333-(14^2) = 137.
338-(14^2) = 142.

no number has a square root in that range.

333-(13^2) = 164.
338-(13^2) = 169.

13 is the square root of 169, so (13^2) + (13^2) = 338, which again shows us ANSWER CHOICE E

bumpbot · Answer

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If x and y are integers and neither x nor y is divisible by 3, which

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