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Get access to the detailed solutions to the previous years questions asked in IIM IPMAT exam
There are multiple ways of solving these sorts of questions. One method is to look for powers of the term in the numerator that leave a remainder of 1 or -1 when divided by the denominator.
Noting down the powers of 3, 3, 9, 27, 81, 243
243 is one such number, 242 is multiple of 11 (11 times 22), hence 243 will leave a remainder of 1 when divided by 11.
243 is 3 raised to power 5; we can rewrite the given term as