The remainder when 5^{99} is divided by 13 is | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) We need to find the remainder of $5^{99}$ when divided by $13$.Using Fermat's Little Theorem,since $13$ is a prime number and $\gcd(5, 13) = 1$,we

Vedclass · Accepted Answer

$8$

Vedclass · Answer

(B) We need to find the remainder of $5^{99}$ when divided by $13$.Using Fermat's Little Theorem,since $13$ is a prime number and $\gcd(5, 13) = 1$,we have $5^{13-1} \equiv 1 \pmod{13}$,which means $5^{12} \equiv 1 \pmod{13}$.We can write $5^{99} = 5^{12 	imes 8 + 3} = (5^{12})^8 	imes 5^3$.Since $5^{12} \equiv 1 \pmod{13}$,then $(5^{12})^8 \equiv 1^8 \equiv 1 \pmod{13}$.Thus,$5^{99} \equiv 1 	imes 5^3 \pmod{13}$.$5^3 = 125$.Dividing $125$ by $13$: $125 = 13 	imes 9 + 8$.Therefore,$125 \equiv 8 \pmod{13}$.The remainder is $8$.

The remainder when $5^{99}$ is divided by $13$ is

Similar Questions

If the greatest divisor of $30 \cdot 5^{2n} + 4 \cdot 2^{3n}$ is $p, \forall n \in N$ and that of $2^{2n+1} - 6n - 2$ is $q, \forall n \in N$, then $p+q=$

The least remainder when $17^{30}$ is divided by $5$ is

The sum of the squares of three consecutive odd numbers increased by $1$ is divisible by-

If $11^{12}-11^2=k(5 \times 10^9+6 \times 10^9+33 \times 10^8+110 \times 10^7+\ldots+33)$,then $k=$

If $P(n):$ " $2^{2n}-1$ is divisible by $k$ for all $n \in N$ " is true,then the value of $k$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module