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.
$1$. The remainder will be the unit digit of the number you are dividing. For example, Remainder when $16$ is divided by $10$ is $6$.
Proof: If you have got a $n$ digit number then you can write it as $10^{n-1}a_0+10^{n-2}a_1+........+10a_{n-2}+a_{n-1}$ where $a_{n-1}$ is the unit digit. Notice that all the terms in the sum are divisible by $10$, the only suspect is $a_{n-1}$.
$2$. Notice that unit digit of a square number can be $0,1,4,5,6,9$ and corresponding unit digits of $4th$ powers can be $0,1,6,5,6,1,$.
So, largest remainder is $6$.
Best Answer
Here is some motivation for the choice of $7$ as the modulus, as you asked. The equation that you want to show that has no solutions in the integers is $$6n^3 +3 -m^6=0.$$ When it comes to polynomial Diophantine equations, especially of the olympiad variety, a common trick is to take everything to one side, look at the equation in a certain modulus $q,$ substitute in all possible combinations of the residues and show that the expression never equals the zero residue. Both because you want to be efficient in your computations and because you want to reduce the chances of everything cancelling out to zero (this heuristic is not rigorous), the idea is to pick a modulus where the various terms in the expression will take on very few distinct values.
As far as I know, there is no general known method of finding the ideal modulus, but there are two general techniques of which I am aware: take advantage of Sophie Germain primes and Fermat's little theorem. Sophie Germain primes $p$ satisfy the fact that $2p+1$ is also a prime, and $3$ is a such a prime. By Fermat's little theorem, if $p$ is a Sophie Germain prime, then $$x^{2p}\equiv 1 \pmod{2p+1}$$ or $x\equiv 0\pmod{2p+1}.$ So $$x^p\equiv \pm 1 \pmod{2p+1}$$ or $x\equiv 0\pmod{2p+1}.$ This means $7$ is a really nice modulus because you have a cube whose resides can only be $0,1,-1,$ and a sixth power whose residues can only be $0,1.$ Then just compute the $2\cdot 3=6$ cases and none will work out.
By the way, years ago I asked the general question on MathOverflow in this thread. (Sadly, I deleted the email address associated with that account and so can no longer access the account, sigh.)