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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
65%
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Question Stats:
62% (02:24) correct
38%
(02:36)
wrong
based on 173
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If y^2 is the sum of the squares of 11 consecutive integers, what is the smallest possible value of the positive integer y?
A. 8
B. 9
C. 10
D. 11
E. 12
A. 8
B. 9
C. 10
D. 11
E. 12
\(y^2\) is a perfect square which is the sum of the squares of \(11\) consecutive numbers. Find the minimum possible value for the positive integer \(y\).
Updated on: 30 Jan 2024, 18:17
Originally posted by crimson_king on 30 Jan 2024, 07:24.
Last edited by crimson_king on 30 Jan 2024, 18:17, edited 3 times in total.
Last edited by crimson_king on 30 Jan 2024, 18:17, edited 3 times in total.
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The answer to this question is Option (D)
We can verify that through the formula for the sum of consecutive squares which is n(n+1)(2n+1)/6. We have to also consider the fact that negative numbers might be involved. We can take the first 5 positive & negative integers along with zero for the lowest possible value of the sum of the 11 squares.
Keeping this in mind the sum of the first 5 positive squares is 55. Since the sum of the first 5 negative squares will also be equal to 55, the least possible value of the sum of the 11 squares will be 55+55+0 (Square of 0)=110. So by this logic we can eliminate options A,B & C as their squares are less than 110. Now for the other two options we can see that if we remove 5 from this list of 11 consecutive integers & put -6 in its place or remove -5 from the list & replace it with 6, the sum of the squares comes out to be 110-25+36=121
So the value comes out to be 121 which is the square of 11, hence option (D) 11 is the correct answer
Posted from my mobile device
We can verify that through the formula for the sum of consecutive squares which is n(n+1)(2n+1)/6. We have to also consider the fact that negative numbers might be involved. We can take the first 5 positive & negative integers along with zero for the lowest possible value of the sum of the 11 squares.
Keeping this in mind the sum of the first 5 positive squares is 55. Since the sum of the first 5 negative squares will also be equal to 55, the least possible value of the sum of the 11 squares will be 55+55+0 (Square of 0)=110. So by this logic we can eliminate options A,B & C as their squares are less than 110. Now for the other two options we can see that if we remove 5 from this list of 11 consecutive integers & put -6 in its place or remove -5 from the list & replace it with 6, the sum of the squares comes out to be 110-25+36=121
So the value comes out to be 121 which is the square of 11, hence option (D) 11 is the correct answer
Posted from my mobile device
30 Jan 2024, 11:10
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Option D should be the one
It is said that \(y^{2}\) is the sum of squares of 11 consecutive integers.
Let's take the first integer as n. So the sequence of 11 consecutive integers will be n,n+1,n+2,n+3,n+4,n+5,n+6,n+7,n+8,n+9,n+10.
S = \(1^{2}\)+\(2^{2}\)+\(3^{2}\)+\(4^{2}\)+\(5^{2}\)+\(6^{2}\)+\(7^{2}\)+\(8^{2}\)+\(9^{2}\)+\(10^{2}\) = n(n+1)(2*n+1)/6 where n=10
so, S= 10*11*21/6 = 35*11
Sum of squares of these numbers is = 11\(n^{2}\) + 2*(n+2n+3n+4n+5n+6n+7n+8n+9n+10n) + 11*35 => 11\(n^{2}\) + 110n +11*35 = 11k
So, the square number is a multiple of 11, which is only \(11^{2}\) among the answer options.
Hence, correct answer should be option D
It is said that \(y^{2}\) is the sum of squares of 11 consecutive integers.
Let's take the first integer as n. So the sequence of 11 consecutive integers will be n,n+1,n+2,n+3,n+4,n+5,n+6,n+7,n+8,n+9,n+10.
S = \(1^{2}\)+\(2^{2}\)+\(3^{2}\)+\(4^{2}\)+\(5^{2}\)+\(6^{2}\)+\(7^{2}\)+\(8^{2}\)+\(9^{2}\)+\(10^{2}\) = n(n+1)(2*n+1)/6 where n=10
so, S= 10*11*21/6 = 35*11
Sum of squares of these numbers is = 11\(n^{2}\) + 2*(n+2n+3n+4n+5n+6n+7n+8n+9n+10n) + 11*35 => 11\(n^{2}\) + 110n +11*35 = 11k
So, the square number is a multiple of 11, which is only \(11^{2}\) among the answer options.
Hence, correct answer should be option D
