You are stepping into modular arithmetic. Basically you work with the remainders after division by something. For example, if you work $\pmod 6$ the possible remainders are $0,1,2,3,4,5$. You can probably convince yourself that you can add, subtract, and multiply any example of numbers and take the remainder before or after the operation and get the same result. That is, if $k=6m+a, l=6n+b$, the remainder of $k+l$ on division by $6$ is the same as the remainder of $k+l$ on division by $6$ and similarly for the other operations. We write that $k+l \equiv a+b \pmod 6$
You are being asked to solve
$n \equiv 9 \pmod {10} \\ n\equiv 8 \pmod 9 \\ n \equiv 7 \pmod 8$
and so on. The Chinese remainder theorem (CRT) guarantees a solution when the moduli are coprime (which they are not here) and says the solutions (if they exist) recur at the least common multiple of the moduli. For this problem there is a trick. You could notice that the equations are the same as
$n \equiv -1 \pmod {10} \\ n\equiv -1 \pmod 9 \\ n \equiv -1 \pmod 8$
and so on. You might notice that one solution is $-1$, so you can find the least common multiple of $2,3,4,\ldots 10$, subtract $1,$ and there you are. This wouldn't work if the remainders didn't have this nice pattern. If they were "random numbers" and the moduli were large you would have to use the full power of the theorem to find the answer.
If the remainders were "random numbers" but the moduli were small, you could also use the CRT in the approach you describe. First find an answer for moduli $2$ and $3$. It isn't hard to see that $5$ is a solution. Then the solutions recur at LCM$(2,3)=6$ so your candidates are $5,11,17,\ldots$ Now look for a solution with modulo $4$ added in and find $11$. Now the solutions recur with period $12$, so they are $11,23,35,\ldots$ Add modulo $5$ into the mix and find $59$. Each time, you only have to look at the number of the modulus or less possibilities and you will get there pretty quickly.
Hint: $6\cdot2=12$, and when $12$ is divided by $7$, the remainder is $5$. What you said is true in general if you take the remainder once again, to ensure that the final answer is less than what you divide by.
See Modular Arithmetic for more information.
Best Answer
$12^{10}=(2^2\times3)^{10} =2^{20}\times 3^{10}$
$m=4k+3$ for some integer $k$ so $22m=88k+66 = 8(11k+8)+2$