2016 AIME II Problem 9

Question

The sequences of positive integers

1,

a_2,

a_3,

\ldots

and

1,

b_2,

b_3,

\ldots

are an increasing arithmetic sequence and an increasing geometric sequence, respectively. Let

c_n = a_n + b_n.

There is an integer

k

such that

c_{k-1} = 100

and

c_{k+1} = 1000.

Find

c_k.

Answer

Answer: 262

Concepts:arithmetic sequence geometric sequence modular arithmetic casework

Difficulty rating: 2920

Small hint:

Write

a_n = 1 + (n-1)d

and

b_n = r^{n-1}

with integers

d \ge 1,

r \ge 2,

and turn the two given values into equations

Big hint:

Show

d

and then

r

must be multiples of

3;

with

r^k \lt 999

only six pairs

(r, k)

remain to test

Solution:

Write

a_n = 1 + (n-1)d

and

b_n = r^{n-1}

with integers

d \ge 1

and

r \ge 2.

Since

c_1 = 2 \lt 100,

we have

k \ge 3,

and the two conditions read

(k-2)d + r^{k-2} = 99,

kd + r^k = 999.

Subtracting,

2d + r^{k-3}(r-1)r(r+1) = 900.

The product of three consecutive integers is divisible by

3,

so

3

divides

2d,

hence

3

divides

d.

Then

(k-2)d + r^{k-2} = 99

forces

3

to divide

r^{k-2},

so

3

divides

r.

The bounds

r^{k-2} \le 98

and

r^k \le 998

leave only

(r, k) = (3, 3),

(3, 4),

(3, 5),

(3, 6),

(6, 3),

(9, 3).

Testing each against

(k-2)d = 99 - r^{k-2}

and

kd = 999 - r^k,

only

(r, k) = (9, 3)

gives a consistent value,

d = 90.

Then

c_3 = 1 + 2 \cdot 90 + 9^2 = 262.

2016 AIME II Problem 9

Answer: 262

Difficulty rating: 2920

Small hint:

Big hint:

Solution:

Problem 9 in other years