2009 AMC 10A Problem 16

Attempt Problem 16 of the 2009 AMC 10A below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2009 AMC 10A solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

16.

Let a,a, b,b, c,c, and dd be real numbers with ∣a−b∣=2,|a - b| = 2, ∣b−c∣=3,|b - c| = 3, and ∣c−d∣=4.|c - d| = 4. What is the sum of all possible values of ∣a−d∣?|a - d|?
99
1212
1515
1818
2424
Answer: D
Concepts:absolute valuecasework
Small hint:
Write a−da - d =(a−b)+(b−c)+(c−d)= (a - b) + (b - c) + (c - d) =±2±3±4= \pm 2 \pm 3 \pm 4
Big hint:
List the distinct values of ∣±2±3±4∣|\pm 2 \pm 3 \pm 4|
Solution:
Since a−da - d =(a−b)+(b−c)+(c−d)= (a-b) + (b-c) + (c-d) =±2±3±4,= \pm 2 \pm 3 \pm 4, the possible absolute values are 2+3+4=9,2+3−4=1,2−3+4=3,−2+3+4=5. \begin{aligned} 2+3+4 &= 9, \\ 2+3-4 &= 1, \\ 2-3+4 &= 3, \\ -2+3+4 &= 5. \end{aligned}
Their sum is 9+1+3+5=18.9 + 1 + 3 + 5 = 18.
Thus, the correct answer is D.
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