AMC 10 Daily Practice - Sequences

Complete problem set with solutions and individual problem pages

Problem 2 Medium

Consider sequences of positive integers for which both the following conditions are true:

(a) each term after the second term is the sum of the two preceding terms;

(b) the eighth term is 260.

How many such sequences are there?

  • A.

    1

  • B.

    2

  • C.

    3

  • D.

    4

  • E.

    5

Answer:B

Let a and b be the first and second terms respectively. Then the first eight terms are

a,b,a+b,a+2b,2a+3b,3a+5b,5a+8b,8a+13b

Hence we seek solutions to the equation

8a+13b=260,  (1)

where a and b are positive integers, since any such solution will generate a sequence of positive integers of the required sort.

Now 8a=13(20-13b=260-b), which is a multiple of 13. Therefore 8a is a multiple of 13, so that a is a multiple of 13.

Let a = 13k, where k is a positive integer. Then equation (1) becomes

8 \times 13k+13k=260 ,

so that

8k+b=20

Since b and k are positive integers there are therefore only two possible values for k, namely 1 and 2. When k=1, we have a =13 and b=12. When k=2, we have a = 26 and b = 4. Hence, there are two possible sequences.