Let N = 53^{91} - 31^{82}. When the value of N is divided by 11, which of the following re… — State PSC Exam

Let $N = 53^{91} - 31^{82}$ . When the value of $N$ is divided by $11$ , which of the following represents the correct remainder?

The correct remainder is 5.

By finding the remainders of the bases (

53 \equiv -2 \pmod{11}

and

31 \equiv -2 \pmod{11}

) and reducing the exponents using Fermat's Little Theorem modulo 10 (

91 \equiv 1 \pmod{10}

and

82 \equiv 2 \pmod{10}

), the expression simplifies to

(-2)^1 - (-2)^2 = -6

. Converting this negative remainder to a true positive remainder yields

-6 + 11 = 5

Determine the remainders of the bases 53 and 31 when divided by 11.

53 \equiv 9 \equiv -2 \pmod{11}

and

31 \equiv 9 \equiv -2 \pmod{11}

Converting to negative remainders significantly simplifies the calculations for large powers.

Apply Fermat's Little Theorem to reduce the exponents.

Since 11 is a prime number, the cyclicity is

11 - 1 = 10

. Reducing the exponents modulo 10 gives

91 \equiv 1 \pmod{10}

and

82 \equiv 2 \pmod{10}

Fermat's theorem states that

a^{p-1} \equiv 1 \pmod{p}

, meaning exponents can be reduced by finding their remainder when divided by

p-1

Evaluate the reduced expression under modulo 11.

(-2)^1 - (-2)^2 = -2 - 4 = -6

Substituting the simplified bases and exponents provides the intermediate remainder.

Convert the negative intermediate remainder into a valid positive remainder.

-6 + 11 = 5

A final remainder must always be a non-negative integer strictly less than the divisor.

Applying Fermat's Little Theorem and correctly handling negative remainders in modular arithmetic.

Soru