This discussion board is a place to discuss our Daily Challenges and the math and science
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explain the steps and thinking strategies that you used to obtain the solution. Comments
should further the discussion of math and science.
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Math
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Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3
2×3
2^{34}
234
a_{i-1}
ai−1
\frac{2}{3}
32
\sqrt{2}
2
\sum_{i=1}^3
∑i=13
\sin \theta
sinθ
\boxed{123}
123
Comments
Suppose that the numbers are x1,x2,…,x10, and suppose that the sum of all the numbers except the jth is the square yj2. If N=x1+x2+⋯+x10, we deduce that
xj=N−yj21≤j≤10
For this to be consistent, we need
N=x1+x2+⋯+x10=10N−(y12+y22+⋯+y102)
so that
y12+y22+⋯+y102=9N
We need to choose y1,y2,⋯,y10 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
x1,x2,…,x10=57,52,45,36,25,12,−3,−20,−60,−83
There is no rule that says that the integers have to be positive.
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??
Easy Math Editor
This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.
When posting on Brilliant:
*italics*
or_italics_
**bold**
or__bold__
paragraph 1
paragraph 2
[example link](https://brilliant.org)
> This is a quote
\(
...\)
or\[
...\]
to ensure proper formatting.2 \times 3
2^{34}
a_{i-1}
\frac{2}{3}
\sqrt{2}
\sum_{i=1}^3
\sin \theta
\boxed{123}
Comments
Suppose that the numbers are x1,x2,…,x10, and suppose that the sum of all the numbers except the jth is the square yj2. If N=x1+x2+⋯+x10, we deduce that xj=N−yj21≤j≤10 For this to be consistent, we need N=x1+x2+⋯+x10=10N−(y12+y22+⋯+y102) so that y12+y22+⋯+y102=9N We need to choose y1,y2,⋯,y10 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 x1,x2,…,x10=57,52,45,36,25,12,−3,−20,−60,−83 There is no rule that says that the integers have to be positive.
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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??
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Not really. There are lots of options. I could have chosen 0,9,18,27,…, but I wanted small numbers.
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9.
Well, I played with the values of squares moduloLog in to reply
Wow..thanks a lot.
This may be simple or difficult ..i tried for a while...need help!
Why graph of quadratic equation is parabola ?