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# Help needed!

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 ago

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happy bday · 2 years ago

Thnx a lot. Actually it was yesterday. · 2 years ago

consider a>b>c>d, then plug in to open mod. then solve by cases. ans is 1,98,99,100. · 2 years ago

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. · 2 years ago

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). · 2 years ago

where did you get the question from . actually you need not solve 24 cases.most of them are same · 2 years ago

Oh... That sounds great I'll try that. My teacher gave me this one. · 2 years ago

Is the answer 98,99,100??? · 2 years ago

Don't know that.

Can u state an example for each of the values? It would be really helpful. · 2 years ago