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PrintSouth African Mathematics Olympiad First Round
South Africa algebra
Problem
A sequence of terms is constructed as follows: The first two terms of the sequence are both equal to . Starting from the third term, each subsequent term is the sum of the preceding two terms. Each of the terms of this sequence , , , , is now divided by and the remainders are added. What is the sum of all the remainders of the terms? (A) (B) (C) (D) (E)
Solution
The first six terms of the sequence are , , , , , , , which are Odd, Odd, Even, Odd, Odd, Even, , since Odd Odd Even and Odd Even Odd. The pattern continues in the same way in cycles of length , with two odd numbers and one even number in each cycle. After division by , the remainders in each cycle are , , , so the sum of the three remainders is . The sum of the first remainders is therefore .
Final answer
C
Techniques
Recurrence relationsModular ArithmeticIntegers