CAT 2024 Slot 2 · Quant · Question 1

Remainders of powers of 3 modulo 11 repeat, so the first job is to find how long that cycle is.

Running through the powers, {3}^{1}\equiv 3, {3}^{2}\equiv 9, {3}^{3}\equiv 27\equiv 5, {3}^{4}\equiv 15\equiv 4 and {3}^{5}\equiv 12\equiv 1 (\mod 11).

The remainder returns to 1 at the fifth power, so the cycle length is 5.

Now reduce the exponent modulo that cycle length: 333=5\times 66+3, so 333\equiv 3 (\mod 5).

Therefore {3}^{333}={\left({3}^{5}\right)}^{66}\times{3}^{3}\equiv{1}^{66}\times{3}^{3}(\mod 11).

That leaves {3}^{3}=27, and 27=2\times 11+5.

The remainder is 5, which is the fourth option.

Shortcut. Since 11 is prime and does not divide 3, Fermat's little theorem gives {3}^{10}\equiv 1 (\mod 11) at once. Then 333=10\times 33+3 leaves {3}^{3}=27\equiv 5, with no cycle hunting at all.