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Areteem Math Challenge II-A: Number Theory
Q9
How did the sum become 2*(3^3+3^(3^2)+3^(3^3)+3^(3^4)) mod 11?
The remainder when dividing by $11$ the first half of the sum ($3^3+3^{3^{2}}+3^{3^{3}}+3^{3^{4}}$) is the same as the remainder when dividing by $11$ the second half of the sum ($3^{3^{5}}+3^{3^{6}}+3^{3^{7}}+3^{3^{8}}$). So, the remainder when dividing by $11$ of the whole sum is the same as the remainder when dividing by $11$ of twice the first half of the sum, that is, $3^3+3^{3^{2}}+3^{3^{3}}+3^{3^{4}}+3^{3^{5}}+3^{3^{6}}+3^{3^{7}}+3^{3^{8}}\equiv 2\cdot (3^3+3^{3^{2}}+3^{3^{3}}+3^{3^{4}})\pmod{11}$.
how did u calculate the (3^3^5+3^3^6+3^3^7+3^3^8) mod 11
Hello again!
You can try finding a pattern in the remainders of powers of $3$ $\pmod{11}$. For instance, $$ \begin{array}{rcl} 3^1 & \equiv & 3 \pmod{11}\\ 3^2 & \equiv & 9 \pmod{11}\\ 3^3 & \equiv & 5 \pmod{11}\\ 3^4 & \equiv & 4 \pmod{11}\\ 3^5 & \equiv & 1 \pmod{11}\\ 3^6 & \equiv & 3 \pmod{11}\\ & \vdots & \end{array} $$ so, you can see that the remainders follow a pattern that repeats every $5$ powers: $$3,9,5,4,1.$$ This way, for example, you can see that $3^{3^5} \equiv 5 \pmod{11}$ since the exponent $3^5$ has remainder $3$ when dividing by $5$.
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