Let the square number y ^ 2 be the sum of 11 consecutive positive integers, and find the minimum value of the positive integer y

Let the sixth of 11 numbers be x, then the first X-5, the second x-4 , 11th x + 5
Then y ^ 2 = (X-5) + (x-4) + +(x+5)=11x
Because 11 is prime, when the minimum x = 11, the right side is a perfect square
Here y = 11

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