How to find the mod of this large number How would one calculate $6000006000600000600006006000000003 \times 3 \pmod{18}$?
The number is too big to use a calculator to manipulate the number into something I can work with, and Fermat's theorem doesn't apply since $18$ isn't prime, so I'm stumped as to what to do.
 A: We can use the Chinese remainder theorem.  Let $(\text{Big Number}) = x$.  Notice that $x \equiv 0 \pmod{9}$ since each factor is divisible by $3$, and further $x \equiv 1 \pmod{2}$ since it is a product of two odd numbers.  We have $\gcd(2, 9) = 1$ and $18 = 2 \times 9$, so the theorem guarantees a solution $\pmod{18}$.  There even exists a method of constructing it explicitly.
Edit: Vadim's answer is pretty good in this case.  The above method will give you a way of attacking similar problems if it isn't as obvious.
A: The digit sum of $600\ldots 03$ is a multiple of $3$, so that monster is a multiple of $3$.  When multiplying by $3$, the result is a multiple of $9$.  However, the result is odd, so modulo $18$, the result is $9$ (not $0$).
A: You imply calculators would be an acceptable means, were it not for the limit on number of digits.  How about:
A: For the specific case, all the $6$s multiplied by $3$ give $18$ because there are enough zeros to make sure the carry isn't a problem.  Since $18 \equiv 0 \pmod {18}$ we can ignore those, so all we care about the the $3 \times 3 =9$  Because the big number is so special, I believe this is the expected solution.  If the big number weren't so special, doing the calculation on paper with long division is not so hard.
A: Let $B = 600000600060000060000600600000000$ (your big number minus $3$). Now the original problem is $3(B+3) \pmod {18}$, or $3B+9 \pmod {18}$
$B$ is divisible by $2$ (last digit is zero) and by $9$ (the sum of the digits is divisible by $9$). Therefore $B$ is divisible by $18$, so your original problem reduces to $3 \cdot 0 + 9 \pmod {18}$, or $ 9 \pmod {18}.$
A: $$6000006000600000600006006000000003\times3=18000018001800001800018018000000009$$ so why worry ?
A: Hint $\bmod \color{#c00}{18}\!:\,\ \color{#c00}3(3\!+\!\color{#c00}6n)\equiv 9$

Or $\ \color{#c00}3(k\!+\!6n)\bmod 18 = \color{#c00}3(k\! +\! 6n\bmod 6) = 3(k\bmod 6)$
employing $\ \color{#c00}ab\bmod \color{#c00}ac = \color{#c00}a(b\bmod c) = $ $\!\bmod\!$ Distributive Law to factor $\,\color{#c00}{a=3}\,$ out of the mod.
A: I looked at it. On the left side we have six times stuff plus six times stuff plus six times stuff plus three more sixes times stuff plus 3, multiplied by 3. 
The result of the multiplication is lots of 18's times stuff, plus 9. 
Modulo 18 is 9. 
Why would you need a calculator, or computer software, or any big theorems? 
