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AMTI Bhaskara Contest Final Junior (October 2013) question no. 8(b)

Do there exist 10 distinct integers such that sum of any 9 of which is a perfect square?

Note by Piyushkumar Palan
4 years ago

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4 votes

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Suppose that the numbers are \(x_1,x_2,\ldots,x_{10}\), and suppose that the sum of all the numbers except the \(j\)th is the square \(y_j^2\). If \(N = x_1+x_2+\cdots+x_{10}\), we deduce that \[ x_j \; = \; N - y_j^2 \qquad 1 \le j \le 10 \] For this to be consistent, we need \[ N \; = \; x_1 + x_2 + \cdots + x_{10} \; = \; 10N - (y_1^2 + y_2^2 + \cdots + y_{10}^2) \] so that \[ y_1^2 + y_2^2 + \cdots + y_{10}^2 \; = \; 9N \] We need to choose \(y_1,y_2,\cdots,y_{10}\) to be distinct integers such that the sum of their squares is divisible by \(9\), for example \(2,3,4,5,6,7,8,9,11,12\). This gives \(N=61\) and \[ x_1,x_2,\ldots,x_{10} \; = \; 57,52,45,36,25,12,-3,-20,-60,-83 \] There is no rule that says that the integers have to be positive.

Mark Hennings - 4 years ago

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Wow..thanks a lot.

Piyushkumar Palan - 4 years ago

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Ahhh....loved it....the method of solving was awesome.......but you did take time to solve the integers \(2,3,4,5,6,7,8,9,11,12\)....right? If not,could you please tell me how you found those integers??

Vaibhav Reddy - 4 years ago

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Not really. There are lots of options. I could have chosen \(0,9,18,27,\ldots\), but I wanted small numbers.

Mark Hennings - 4 years ago

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@Mark Hennings so you arithmetically solved it on a piece of paper

Vaibhav Reddy - 4 years ago

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@Vaibhav Reddy Well, I played with the values of squares modulo \(9\).

Mark Hennings - 4 years ago

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@Mark Hennings Thank you....I realized it later....

Vaibhav Reddy - 4 years ago

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This may be simple or difficult ..i tried for a while...need help!

Piyushkumar Palan - 4 years ago

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Why graph of quadratic equation is parabola ?

Vaibhav Gandhapwad - 2 years, 5 months ago

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