Let S be the set of values which can be written as the sum of 5 consecutive perfect squares.
What is the smallest element of S which is divisible by 17?
Choose the correct choice from the following answers:
A) 255 B) 355 C) 522 D) 355 E) 664
Each element of S can be written in the form   (x- 2)2 + (x - 1)2 + x2 + (x + 1)2 + (x + 2)2 = x2 + 2x2 + 2 + 2x2 + 8   = 5x2 + 10   = 5(x2 + 2)   where x is some positive integer with x 3. By iterating over values of x, we observe that when   x = 7, x2 + 2 = 51 = 3 * 17; no smaller value of x exists which satisfies the problem statement.   Hence, the solution is 5(x2 + 2) = 5 *51= 
<gadget id="calculator">5 *51</gadget>
<output>255</output>
255   correct answer 255
<result>A</result>
Let S be the set of values which can be written as the sum of 5 consecutive perfect squares.
What is the smallest element of S which is divisible by 17?