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Question: Let x1,x2,x3,x4,x5 belong to R and |x1-x2| = 2, |x2-x3| = 4, |x3-x4| = 3, |x4-x5| = 5. Then the sum ...

Let x1,x2,x3,x4,x5 belong to R and |x1-x2| = 2, |x2-x3| = 4, |x3-x4| = 3, |x4-x5| = 5. Then the sum of all distinct possible values of |x5-x1| is

Answer

44

Explanation

Solution

The problem asks for the sum of all distinct possible values of ∣x5−x1∣|x_5 - x_1| given a series of absolute differences between consecutive terms.

Given:

  1. ∣x1−x2∣=2  ⟹  x1−x2=±2|x_1 - x_2| = 2 \implies x_1 - x_2 = \pm 2
  2. ∣x2−x3∣=4  ⟹  x2−x3=±4|x_2 - x_3| = 4 \implies x_2 - x_3 = \pm 4
  3. ∣x3−x4∣=3  ⟹  x3−x4=±3|x_3 - x_4| = 3 \implies x_3 - x_4 = \pm 3
  4. ∣x4−x5∣=5  ⟹  x4−x5=±5|x_4 - x_5| = 5 \implies x_4 - x_5 = \pm 5

We want to find the possible values of ∣x5−x1∣|x_5 - x_1|.
Let's express x5−x1x_5 - x_1 as a sum of the given differences:
x5−x1=(x5−x4)+(x4−x3)+(x3−x2)+(x2−x1)x_5 - x_1 = (x_5 - x_4) + (x_4 - x_3) + (x_3 - x_2) + (x_2 - x_1)

From the given conditions, we can write: x2−x1=a1x_2 - x_1 = a_1, where a1∈{2,−2}a_1 \in \{2, -2\}
x3−x2=a2x_3 - x_2 = a_2, where a2∈{4,−4}a_2 \in \{4, -4\}
x4−x3=a3x_4 - x_3 = a_3, where a3∈{3,−3}a_3 \in \{3, -3\}
x5−x4=a4x_5 - x_4 = a_4, where a4∈{5,−5}a_4 \in \{5, -5\}

So, x5−x1=a4+a3+a2+a1x_5 - x_1 = a_4 + a_3 + a_2 + a_1.
This means x5−x1=(±5)+(±3)+(±4)+(±2)x_5 - x_1 = (\pm 5) + (\pm 3) + (\pm 4) + (\pm 2).

Let S=x5−x1S = x_5 - x_1. We need to find all distinct possible values of SS.
There are 24=162^4 = 16 possible combinations of signs for the terms.

Let's determine the range of SS:
Maximum value of S=5+3+4+2=14S = 5 + 3 + 4 + 2 = 14.
Minimum value of S=−5−3−4−2=−14S = -5 - 3 - 4 - 2 = -14.

All terms 2,4,3,52, 4, 3, 5 are integers, so their sum or difference will always be an integer.
Let's check the parity of SS:
S=(±2)+(±4)+(±3)+(±5)S = (\pm 2) + (\pm 4) + (\pm 3) + (\pm 5)
The terms (±2)(\pm 2) and (±4)(\pm 4) are always even.
The terms (±3)(\pm 3) and (±5)(\pm 5) are always odd.
The sum of two even numbers is even. The sum of two odd numbers is even.
Therefore, S=(even+even)+(odd+odd)=even+even=evenS = (\text{even} + \text{even}) + (\text{odd} + \text{odd}) = \text{even} + \text{even} = \text{even}.
This means all possible values of SS must be even integers.

Let's list all possible values of SS systematically, starting from the maximum value (14) and seeing what values can be obtained by flipping signs.
Let Sall_pos=2+4+3+5=14S_{all\_pos} = 2+4+3+5 = 14.

  1. Flipping one sign:

    • Change 2→−22 \to -2: 14−2(2)=14−4=1014 - 2(2) = 14 - 4 = 10. (e.g., −2+4+3+5=10-2+4+3+5=10)
    • Change 4→−44 \to -4: 14−2(4)=14−8=614 - 2(4) = 14 - 8 = 6. (e.g., 2−4+3+5=62-4+3+5=6)
    • Change 3→−33 \to -3: 14−2(3)=14−6=814 - 2(3) = 14 - 6 = 8. (e.g., 2+4−3+5=82+4-3+5=8)
    • Change 5→−55 \to -5: 14−2(5)=14−10=414 - 2(5) = 14 - 10 = 4. (e.g., 2+4+3−5=42+4+3-5=4)

  2. Flipping two signs:

    • Change 2,4→−2,−42, 4 \to -2, -4: 14−2(2)−2(4)=14−4−8=214 - 2(2) - 2(4) = 14 - 4 - 8 = 2. (e.g., −2−4+3+5=2-2-4+3+5=2)
    • Change 2,3→−2,−32, 3 \to -2, -3: 14−2(2)−2(3)=14−4−6=414 - 2(2) - 2(3) = 14 - 4 - 6 = 4. (already found)
    • Change 2,5→−2,−52, 5 \to -2, -5: 14−2(2)−2(5)=14−4−10=014 - 2(2) - 2(5) = 14 - 4 - 10 = 0. (e.g., −2+4+3−5=0-2+4+3-5=0)
    • Change 4,3→−4,−34, 3 \to -4, -3: 14−2(4)−2(3)=14−8−6=014 - 2(4) - 2(3) = 14 - 8 - 6 = 0. (already found)
    • Change 4,5→−4,−54, 5 \to -4, -5: 14−2(4)−2(5)=14−8−10=−414 - 2(4) - 2(5) = 14 - 8 - 10 = -4. (e.g., 2−4+3−5=−42-4+3-5=-4)
    • Change 3,5→−3,−53, 5 \to -3, -5: 14−2(3)−2(5)=14−6−10=−214 - 2(3) - 2(5) = 14 - 6 - 10 = -2. (e.g., 2+4−3−5=−22+4-3-5=-2)

  3. Flipping three signs:

    • Change 2,4,3→−2,−4,−32, 4, 3 \to -2, -4, -3: 14−2(2)−2(4)−2(3)=14−4−8−6=−414 - 2(2) - 2(4) - 2(3) = 14 - 4 - 8 - 6 = -4. (already found)
    • Change 2,4,5→−2,−4,−52, 4, 5 \to -2, -4, -5: 14−2(2)−2(4)−2(5)=14−4−8−10=−814 - 2(2) - 2(4) - 2(5) = 14 - 4 - 8 - 10 = -8. (e.g., −2−4+3−5=−8-2-4+3-5=-8)
    • Change 2,3,5→−2,−3,−52, 3, 5 \to -2, -3, -5: 14−2(2)−2(3)−2(5)=14−4−6−10=−614 - 2(2) - 2(3) - 2(5) = 14 - 4 - 6 - 10 = -6. (e.g., −2+4−3−5=−6-2+4-3-5=-6)
    • Change 4,3,5→−4,−3,−54, 3, 5 \to -4, -3, -5: 14−2(4)−2(3)−2(5)=14−8−6−10=−1014 - 2(4) - 2(3) - 2(5) = 14 - 8 - 6 - 10 = -10. (e.g., 2−4−3−5=−102-4-3-5=-10)

  4. Flipping four signs:

    • Change 2,4,3,5→−2,−4,−3,−52, 4, 3, 5 \to -2, -4, -3, -5: 14−2(2)−2(4)−2(3)−2(5)=14−4−8−6−10=−1414 - 2(2) - 2(4) - 2(3) - 2(5) = 14 - 4 - 8 - 6 - 10 = -14. (e.g., −2−4−3−5=−14-2-4-3-5=-14)

The set of all distinct possible values for S=x5−x1S = x_5 - x_1 is:
V={14,10,6,8,4,2,0,−4,−2,−8,−6,−10,−14}V = \{14, 10, 6, 8, 4, 2, 0, -4, -2, -8, -6, -10, -14\}.
Arranging them in ascending order:
V={−14,−10,−8,−6,−4,−2,0,2,4,6,8,10,14}V = \{-14, -10, -8, -6, -4, -2, 0, 2, 4, 6, 8, 10, 14\}.

Now, we need to find the distinct possible values of ∣x5−x1∣|x_5 - x_1|, which are ∣S∣|S|.
Taking the absolute value of each element in VV:
∣V∣={∣−14∣,∣−10∣,∣−8∣,∣−6∣,∣−4∣,∣−2∣,∣0∣,∣2∣,∣4∣,∣6∣,∣8∣,∣10∣,∣14∣}|V| = \{|-14|, |-10|, |-8|, |-6|, |-4|, |-2|, |0|, |2|, |4|, |6|, |8|, |10|, |14|\}
∣V∣={14,10,8,6,4,2,0,2,4,6,8,10,14}|V| = \{14, 10, 8, 6, 4, 2, 0, 2, 4, 6, 8, 10, 14\}

The set of distinct possible values of ∣x5−x1∣|x_5 - x_1| is:
D={0,2,4,6,8,10,14}D = \{0, 2, 4, 6, 8, 10, 14\}.

Finally, we need to find the sum of all these distinct values:
Sum =0+2+4+6+8+10+14= 0 + 2 + 4 + 6 + 8 + 10 + 14
Sum =(2+4+6+8+10)+14= (2+4+6+8+10) + 14
Sum =30+14=44= 30 + 14 = 44.

The final answer is 44\boxed{44}.