Problem:
Let a,b,c, and d be real numbers with ∣a−b∣=2,∣b−c∣=3, and ∣c−d∣=4. What is the sum of all possible values of ∣a−d∣?
Answer Choices:
A. 9
B. 12
C. 15
D. 18
E. 24
Solution:
The given conditions imply that b=a±2,c=b±3=a±2±3, and d=c±4=a±2±3±4, where the signs can be combined in all possible ways. Therefore the possible values of ∣a−d∣ are 2+3+4=9,2+3−4=1, 2−3+4=3, and −2+3+4=5. The sum of all possible values of ∣a−d∣ is 9+1+3+5=(D)18​.
OR
The equations in the problem statement are true for numbers a,b,c,d if and only if they are true for a+r,b+r,c+r,d+r, where r is any real number. The value of ∣a−d∣ is also unchanged with this substitution. Therefore there is no loss of generality in letting b=0, and we can then write down the possibilities for the other variables:
The different possible values for ∣a−d∣ are
∣2−7∣=5,∣2−(−1)∣=3,∣2−1∣=1,∣2−(−7)∣=9
The sum of these possible values is (D)18​.
The problems on this page are the property of the MAA's American Mathematics Competitions
