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.....

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Is the answer 98,99,100???

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Don't know that.

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

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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.

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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).

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

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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.

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happy bday

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Thnx a lot. Actually it was yesterday.

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