2005 AIME II Problem 11

Question

Let

m

be a positive integer, and let

a_0,

a_1,

\ldots,

a_m

be a sequence of real numbers such that

a_0 = 37,

a_1 = 72,

a_m = 0,

and

a_{k+1} = a_{k-1} - \frac{3}{a_k}

for

k = 1,

2,

\ldots,

m - 1.

Find

m.

Answer

Multiply the recurrence by

a_k

to see that the products

a_k a_{k-1}

drop by

3

at each step

a_1 a_0 = 2664 = 3 \cdot 888,

and

a_m = 0

exactly when the product

a_m a_{m-1}

reaches

0

Multiplying the recurrence by

a_k

gives

a_{k+1} a_k = a_k a_{k-1} - 3,

so the products

b_k = a_k a_{k-1}

form an arithmetic sequence with common difference

-3.

Since

b_1 = 72 \cdot 37 = 2664 = 3 \cdot 888,

we get

\begin{aligned} b_k &= 2664 - 3(k - 1) \\ &= 3(889 - k). \end{aligned}

Thus

b_k \gt 0

for

k \le 888,

so no term before

a_{889}

can vanish (and the recurrence never divides by zero), while

b_{889} = a_{889} a_{888} = 0

with

a_{888} \ne 0.

Hence

a_{889} = 0,

and

m = 889.

Problem 11 in other years