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Let $a,b,c,d$ real numbers such that $\lvert\,a-b\,\rvert=2,\lvert\,b-c\,\rvert=3,\lvert\,c-d\,\rvert=4$.Then the sum of all possible values of $\lvert\,a-d\,\rvert$ is...

Now obviously one method is to take off the modulus and and take cases... which would obviously be a lengthy method.

Is there any method which would take this sum down in $6$ or $7$ lines...

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  • $\begingroup$ $d$ isn't represented in your absolute differences; it seems completely unrestrained. $\endgroup$
    – Tom
    Oct 24, 2017 at 14:37
  • $\begingroup$ questio has been editted $\endgroup$
    – Pole_Star
    Oct 25, 2017 at 7:46

2 Answers 2

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I'm reading your third condition as $|c-d|=4$.

I'm lining up the points on a figure showing the number line with ticks. WLOG we may assume $a=0$, $b=2$ and then obtain for $c$ the two possibilities $c_1=-1$, $c_2=5$. This leads to the four possible $d$-values $d_{11}=-5$, $d_{12}=3$, $d_{21}=1$, $d_{22}=9$. The sum of the absolute values of these four numbers is $18$.

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| a – b | = 2
⇒ a – b = ± 2
| a – c | = 3
⇒ b – c = ± 3
| c – d | = 4 ⇒ c – d = ± 4 Possible values of a – d are ± 9, ± 5, ± 3, ± 1 | a – d | = 9, 5, 3, 1
Sum = 18

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