What is $2^{-1} + 2^{-2} + 2^{-3} + 2^{-4} + 2^{-5} + 2^{-6} \pmod{13}$? Express your answer as an integer from $0$ to $12$, inclusive.

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Rod · Answer 1 · 2017-12-10T15:50:55+0000

S=1/2+1/4+1/8+...+1/8192.

S/2=1/4+1/8+1/16+...+1/8192+1/16384.

S-S/2=S/2=1/2-1/16384=8192/16384-1/16384=8191/16384.

So S=2×8191/16384=8191/8192.

anonymous · Answer 2 · 2019-06-01T03:37:01+0000

Let the given sum be $S$ . By inspection, we find that $2^6 \equiv 64 \equiv -1 \pmod{13}$ , so $2^{-6} \equiv (-1)^{-1} \equiv -1 \pmod{13}$ . It follows that $2^{-5} \equiv 2 \cdot 2^{-6} \equiv 2 \cdot -1 \equiv -2 \pmod{13}$ , and that $2^{-4} \equiv -4 \pmod{13}$ , and so forth. Thus, $S \equiv -2^5 - 2^4 - 2^3 - 2^2 - 2 - 1 \equiv -63 \equiv \boxed{2} \pmod{13}$

Solution 2:

Let the given sum be $S$ . The derivation of the geometric series formula remains the same if we consider the result $\mod{13}$ . In particular, $\begin{align*}S &\equiv 2^{-1} \cdot (1 + 2^{-1} + \cdots + 2^{-5}) \pmod{13} \\ 2^{-1} S &\equiv 2^{-1} \cdot (2^{-1} + \cdots + 2^{-5} + 2^{-6}) \pmod{13}.\end{align*}$ Subtracting the two yields that $(1 - 2^{-1}) \cdot S \equiv 2^{-1} \cdot (1 - 2^{-6}) \pmod{13}.$ Since $2 \cdot (1-2^{-1}) \equiv 2 - 1 \equiv 1 \pmod{13}$ , it follows that $1-2^{-1} \equiv 2^{-1} \pmod{13}$ , so $S \equiv 2^{-1} \cdot \frac{1 - 2^{-6}}{1-2^{-1}} \equiv 1 - 2^{-6} \equiv \frac{63}{64} \equiv \frac{-2}{-1} \equiv \boxed{2} \pmod{13}.$ *AoPS answer

anonymous · Answer 3 · 2022-02-08T01:15:21+0000

Let the given sum be $S$ . By inspection, we find that $2^6 \equiv 64 \equiv -1 \pmod{13}$ , so $2^{-6} \equiv (-1)^{-1} \equiv -1 \pmod{13}$ . It follows that $2^{-5} \equiv 2 \cdot 2^{-6} \equiv 2 \cdot -1 \equiv -2 \pmod{13}$ , and that $2^{-4} \equiv -4 \pmod{13}$ , and so forth. Thus, $S \equiv -2^5 - 2^4 - 2^3 - 2^2 - 2 - 1 \equiv -63 \equiv \boxed{2} \pmod{13}$

Solution 2:

Let the given sum be $S$ . The derivation of the geometric series formula remains the same if we consider the result $\mod{13}$ . In particular, $\begin{align*}S &\equiv 2^{-1} \cdot (1 + 2^{-1} + \cdots + 2^{-5}) \pmod{13} \\ 2^{-1} S &\equiv 2^{-1} \cdot (2^{-1} + \cdots + 2^{-5} + 2^{-6}) \pmod{13}.\end{align*}$ Subtracting the two yields that $(1 - 2^{-1}) \cdot S \equiv 2^{-1} \cdot (1 - 2^{-6}) \pmod{13}.$ Since $2 \cdot (1-2^{-1}) \equiv 2 - 1 \equiv 1 \pmod{13}$ , it follows that $1-2^{-1} \equiv 2^{-1} \pmod{13}$ , so $S \equiv 2^{-1} \cdot \frac{1 - 2^{-6}}{1-2^{-1}} \equiv 1 - 2^{-6} \equiv \frac{63}{64} \equiv \frac{-2}{-1} \equiv \boxed{2} \pmod{13}.$

anonymous · Answer 4 · 2022-07-28T13:18:09+0000

2^6 ≡ 64 ≡ -1 (mod 13),

2^(-6) ≡ (-1)^(-1) ≡ -1 (mod 13),

2^(-5) ≡ 2 • 2^(-6) ≡ 2 • -1 ≡ -2 (mod 13),

2^(-4) ≡ 2^2 • 2^(-6) ≡ 4 • -1 ≡ -4 (mod 13).

Finally, - 2^5 - 2^4 - 2^3 - 2^2 - 2^1 - 2^0 ≡ -63 ≡ 2 (mod 13). (So it’s 2.)

What is $2^{-1} + 2^{-2} + 2^{-3} + 2^{-4} + 2^{-5} + 2^{-6} \pmod{13}$? Express your answer as an integer from $0$ to $12$, inclusive.

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