What is the remainder when $213987654213473846989272654857367287454572836418486364$ is divided by $48$? Can it be done by hand i.e. to find the remainder when  $213987654213473846989272654857367287454572836418486364$ is divided by $48$?
 A: Let the large integer $N$. Since $48 = 2^4 \cdot 3$, by the Chinese Remainder Theorem, it suffices to find the remainder $N$ when is divided by $16$ and $3$. 
Since $10^4$ is divisible by $16 = 2^4$, we only need to look at the last four digits to determine the remainder when $N$ is divided by $16$. 
We can look at the sum of the digits of $N$ mod $3$ to get the remainder when $N$ is divided by $3$. 
Both of these are easy to do by hand. 
A: Hint $\ \begin{align} n\equiv a&\!\!\!\pmod 3\\ n\equiv b&\!\!\!\pmod {\color{#c00}{16}}\end{align}\!\!\!\!\overset{\ \ \rm CRT}\iff n\equiv b+16(a\!-\!b)\pmod{48},\ $ and $\ 2\mid 10\,\Rightarrow\,\color{#c00}{2^4}\!\mid 10^4,\,$
and, $\ {\rm mod}\ 3\!:\ 10\equiv 1\,\Rightarrow\, n = f(10)\equiv f(1)\equiv $ digit sum (casting out threes, like nines).
A: Let the giant number be $N$, thus sum of the digits is $276$ (yes, I calculated that by hand) and $48 = 2^4 \times 3$. 
Now sum of digits is $276$, which tells us that it gives $0$ as remainder when divided by $3$ as $3$ divides $276$, so $N \bmod 3 = 0$.
Last four digits are  $6364 = 16 \times 397 + 12$ (hardly five minutes, without calculator, I took around two).
Now $(16, 3) = 1$. Therefore Chinese remainder theorem tells us we need to find the number between $0$ and $48$ such that it is divisible by $3$ and leaves remainder $12$ when divided by $16$.
Only $12$ and $44$ leaves remainder $12$ when divided by $16$, but $44$ is not divisible by $3$.
Hence, the remainder is $12$.
A: One way would be to perform a long division by $48$, which is not even that hard to do.
Another way is to compute the remainder modulo $3$, which can be obtained from the digit sum, and the remainder modulo $16$, which is determined only by the last four digits. 
