AMC 10 Daily Practice - Sequences

,

then .

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

Hence we seek solutions to the equation

,  (1)

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

Now , which is a multiple of . Therefore is a multiple of , so that a is a multiple of .

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

so that

Since and are positive integers there are therefore only two possible values for , namely and . When , we have and . When , we have and . Hence, there are two possible sequences.

The condition where gives the motivation to make a substitution to change the equilibrium from to . We can substitute to achieve that. Now, we need to find the smallest value of such that given that .

Factoring the recursion , we get:

.

Using wishful thinking, we can simplify the recursion as follows:

.

The recursion looks like a geometric sequence with the ratio changing slightly after each term. Notice from the recursion that the sequence is strictly decreasing, so all the terms after will be less than 1. Also, notice that all the terms in sequence will be positive. Both of these can be proven by induction.

With both of those observations in mind, . Combining this with the fact that the recursion resembles a geometric sequence, we conclude that

is approximately equal to and the ranges that the answer choices give us are generous, so we should use either or to find a rough estimate for .

Since , that means . Additionally,

Therefore, we can estimate that .

Raising both sides to the 40th power, we get

But , so and therefore, .

This tells us that is somewhere around 120, so our answer is .