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If \(a, b, c, d\in\mathbb{R}\) such that

\[|a-b|+|c-d|=99\]

\[|a-c|+|b-d|=1\]

then the possible value(s) of \(|a-d|+|b-c|\) are.....

Note by Aneesh Kundu
2 years, 11 months ago

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happy bday

Rajat Kharbanda - 2 years, 11 months ago

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Thnx a lot. Actually it was yesterday.

Aneesh Kundu - 2 years, 11 months ago

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consider a>b>c>d, then plug in to open mod. then solve by cases. ans is 1,98,99,100.

Rajat Kharbanda - 2 years, 11 months ago

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I was able to prove that \[0\leq |a-d|+|b-c|\leq 100\]

I coudn't go any further. I have not verified the equality cases either.

Aneesh Kundu - 2 years, 11 months ago

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If \(a>b>c>d\) then we can remove the modulus from both the equations. Rearranging the last equation we get our expression, but that would give us the value to be 1.

Cases is not really a gud option (24 cases is just tooooo much).

Aneesh Kundu - 2 years, 11 months ago

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where did you get the question from . actually you need not solve 24 cases.most of them are same

Rajat Kharbanda - 2 years, 11 months ago

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@Rajat Kharbanda Oh... That sounds great I'll try that. My teacher gave me this one.

Aneesh Kundu - 2 years, 11 months ago

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Is the answer 98,99,100???

Abhineet Nayyar - 2 years, 11 months ago

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Don't know that.

Can u state an example for each of the values? It would be really helpful.

Aneesh Kundu - 2 years, 11 months ago

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