Problem:
Let a,b,c, and d be real numbers with β£aβbβ£=2,β£bβcβ£=3, and β£cβdβ£=4. What is the sum of all possible values of β£aβdβ£?
Answer Choices:
A. 9
B. 12
C. 15
D. 18
E. 24
Solution:
The given conditions imply that b=aΒ±2,c=bΒ±3=aΒ±2Β±3, and d=cΒ±4=aΒ±2Β±3Β±4, where the signs can be combined in all possible ways. Therefore the possible values of β£aβdβ£ are 2+3+4=9,2+3β4=1, 2β3+4=3, and β2+3+4=5. The sum of all possible values of β£aβdβ£ is 9+1+3+5=(D)18β.
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 β£aβdβ£ 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 β£aβdβ£ are
β£2β7β£=5,β£2β(β1)β£=3,β£2β1β£=1,β£2β(β7)β£=9
The sum of these possible values is (D)18β.
The problems on this page are the property of the MAA's American Mathematics Competitions