Divisibility Chains

We see that for $i=1\to n$ , we have $b_i\equiv 0\pmod{a_i}$ . However, we can also describe these relations using only the variables $a_1$ and $b_1$ .

We see that $b_1\equiv 0\pmod{a_1}$

Next, we have $b_1+1\equiv 0\pmod{a_1+1}\implies b_1\equiv a_1\pmod{a_1+1}$

Next, we have $b_1+2\equiv 0\pmod{a_1+2}\implies b_1\equiv a_1\pmod{a_1+2}$

Generally, we have $b_1\equiv a_1\pmod{a_1+k}$ for $k=0\to n-1$ .

Thus, $b_1=a_1+\text{lcm}(a_1,a_1+1,\ldots ,a_1+n-1)$

To maximize $n$ , we must minimize $a_1$ in order to make sure $b_1\le 1000$ .

Thus, we set $a_1=2$ . Thus, we want $2+\text{lcm}(2,3,\ldots ,n+1)\le 1000$

Checking successive numbers for $n$ , we see that $n=7$ is the highest we can go such that $2+\text{lcm}(2,3,\ldots ,n+1)\le 1000$ .

Thus, $b_1=2+\text{lcm}(2,3,\ldots ,8)=842$ and $b_7=848$ . Also, since $a_1=2$ we have $a_7=8$ . Finally, we have $k=7$ . Thus, the answer is $7+8+848=\boxed{863}$ .

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Checking, we do indeed have the following: $2\mid 842$ $3\mid 843$ $4\mid 844$ $5\mid 845$ $6\mid 846$ $7\mid 847$ $8\mid 848$