Question

The remainder obtained when$\left({104}^{303}\right)$ is divided by 101 is

• A. 3
• B. 27
• C. 55
• D. 98

Answer 1

The correct option is B

27

${104}^{303}$ =${104}^{101}\times {104}^{101}\times {104}^{101}$

According to Fermat`s little theorem (n should be a prime number), when ${104}^{101}$ is divided by 101 the remainder is 3.

Remainder of a product when it is being divided is going to be the same as the product of the individual remainders.

Rem$({104}^{101}\times {104}^{101}\times {104}^{101})$ = Rem$(3\times 3\times 3)$ = 27 is the remainder