<s>[INST] Combinatorics error correcting code (56) * (36)^4 * (-55) + (35)(67)(-14)^2 mod 17. Find the least non-negative residue of the expression module the given n. 
First I just want to make sure I understand what the question wants. To do that, I give a simple case 15 mod 7. 1 is what we looking for right?
Then get back to the question, what is the method for effecticiently finding the answer other than just use calculator to multiply everything and then find the multiple of 7 to minus that member. By the way no use calculator
 [/INST] The calculation can be done without a calculator. Since $56=3\cdot 17+5$, we have $56\equiv 5\pmod{17}$. We have $36\equiv 2\pmod{17}$, so $36^4\equiv 2^4\equiv -1\pmod{17}$. But $(-1)(-55)=55\equiv 4\pmod{17}$. Thus the first part of our expression is $\equiv 20\equiv 3\pmod{17}$. 
A similar calculation shows that the second part is $\equiv 8\pmod{17}$. 
For $35\equiv 1\pmod{17}$, and $67\equiv 16\pmod{17}$. We have $-14\equiv 3\pmod{17}$, so $(-14)^2\equiv 9\pmod{17}$. It remains to calculate $(16)(9)$ modulo $17$. One can work directly, multiplying to get $144$, and finding the remainder. But here is a useful trick: $16\equiv -1\pmod{17}$, so $(16)(9)\equiv (-1)(9)=-9\equiv 8\pmod{17}$. 
Now add. The sum is $\equiv 11\pmod{17}$. 
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