2002 AIME I Problems/Problem 8

Problem

Find the smallest integer $k$ for which the conditions

(1) $a_1,a_2,a_3\cdots$ is a nondecreasing sequence of positive integers

are satisfied by more than one sequence.

Suppose that $a_1=x_0$ is the smallest possible value for $a_1$ that yields a good sequence, and $a_2=y_0$ in this sequence. So, $13x_0+21y_0=k$ .

Since $\gcd(13,21)=1$ , the next smallest possible value for $a_1$ that yields a good sequence is $a_1=x_0+21$ . Then, $a_2=y_0-13$ .

Thus, $k=13(1)+21(35)=\boxed{748}$ is the smallest possible value of $k$ .