elementary number theory - Remainder in Division using divisibility tests

I'm a bit lost / behind in my number theory class, but hey what can I do but try to catch up.

I'm asked to find the remainder in division by $m= 3,7,9,11,13$ using divisbility tests for $1234567 \times 10^{89}$

For division by 3 (and 9), I know that the sum of the digits must be divisible by 3 (or 9).

Immediately it follows that both 3 and 9 do not divide our monster number. To find the remainder, this is what I did though I'm not sure if I am correct.

$28 \equiv 1 \mod 3$

and so I say that the remainder in division by 3 of $1234567 \times 10^{89}$ is 1.

For 7,11,13 I see that
$123 - 456 + 700 = 367$ and that none of 7,11,13 divide 367.

This is where I'm stuck -- the test is developed using the fact that $7*11*13 = 1001$ and $1000 \equiv -1 \mod 1001$.

Do i set $367 \equiv x \mod 143$ ? Or is the remainder in division from 7,11 and 13 all $367 \mod 1001$?