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

The expression ${49^n} + 16n - 1$ is divisible by:

If $(11)^{27} + (21)^{27}$ is divided by $16$,the remainder is:

Among the statements:
$(S1):$ $2023^{2022} - 1999^{2022}$ is divisible by $8$.
$(S2):$ $13(13^{n}) - 11n - 13$ is divisible by $144$ for infinitely many $n \in N$.

For all $n \in N$,if $n(n^2+3)$ is divisible by $k$,then the maximum value of $k$ is

The numbers $a_n = 6^n - 5n$ for $n = 1, 2, 3, \ldots$ when divided by $25$ leave the remainder:

Vedclass Test Series

Exam Paper Generator

Online Exam Module