Over 2000 numbers are around a circle. Each number is the sum of its left and right neighbors.
Given that one of the numbers is a one, how many numbers (as a minimum) must there be?
|
|
Submitted by McWorter
|
|
Rating: 3.7500 (4 votes)
|
|
|
Solution:
|
(Hide)
|
|
Let n be the number of numbers on the circle. Suppose the neighbor clockwise of 1 is a. Then, starting from 1 clockwise the first 8 numbers are
1, a, a-1, -1, -a, 1-a, 1, a.
Since the 7th and 8th numbers are 1 and a, this sequence of 6 numbers repeats itself.
If n is not a multiple of 6, then some two consecutive numbers in the sequence of 6 numbers other than the first two, (1,a), equals (1,a). Upon checking the five possiblilities we find that this is impossible. Hence n is a multiple of 6. We check the five cases.
(1,a)=(a,a-1) implies a=a-1 implies 0=-1; contradiction.
(1,a)=(a-1,-1) implies 1=a-1 and a=-1; implying 1=a-1=-2; contradiction.
(1,a)=(-1,-a) implies 1=-1; contradiction.
(1,a)=(-a,1-a) implies 1=-a and a=1-a, implying a=-1 and a=1/2; contradiction.
(1,a)=(1-a,1) implies 1=1-a and a=1, implying a=0 and a=1; contradiction.
Thus the least admissable value of n greater than 2000 is 2004, the smallest multiple of 6 greater than 2000.
(The number 1 on the circle is not sacred. It merely prohibits the "all zeros solution".) |
