Question

find the remainder when 2^35 is divided by 7​

Answer 1

Answer:

2^35 is divided by 7 is fifth two

Answer 2

Answer:

Remainder is 4

Step-by-step explanation:

Rule1: If $\frac{a^{evenno}}{a+1}$, then the remainder will always be 1.

Rule2: If $\frac{a^{oddno}}{a+1}$, then the remainder will always be a.

Rule3: If $\frac{a^{no}}{a-1}$, then the remainder will always be 1, whether n is even or odd.

$2^{35} = x(mod \ 7)$

$2^{30} * 2^{5} * 2^{1}= x(mod 7)$

$(2^{3})10 * (2^{1})10 * 2^1 = x(mod 7)$

$(8)^{10} * (2) * 2 =x(mod \ 7)$

$(8^2)^{5} * 2* 2= x(mod 7)$

$= 1 * 2 * 2 (mod 7) [(8)^2 = 64 = 1 (mod 7)] = 4 (mod 7)$

$2^{35} = 4(mod 7)$

$x = 4$

The remainder is 4