IPMAT Indore 2025 · Quantitative Ability · Question 40

Handle the two pieces separately, then add the remainders.

Step 1 — reduce the first term’s base.

11\equiv 2 (\mod 9) \Rightarrow{11}^{1011}\equiv{2}^{1011}(\mod 9)

Step 2 — find the cycle of powers of 2 mod 9.

{2}^{1}\equiv 2,{2}^{2}\equiv 4,{2}^{3}\equiv 8,{2}^{4}\equiv 7,{2}^{5}\equiv 5,{2}^{6}\equiv 1 (\mod 9)

The pattern repeats every 6 steps.

1011=6\times 168+3 \Rightarrow{2}^{1011}\equiv{2}^{3}\equiv 8 (\mod 9)

Step 3 — the second term. Use the digit-sum test for 9: 1+0+1+1=3, so

1011\equiv 3 (\mod 9) \Rightarrow{1011}^{11}\equiv{3}^{11}(\mod 9)

But {3}^{2}=9\equiv 0, and 11\ge 2, so {3}^{11} is a multiple of 9:

{1011}^{11}\equiv 0 (\mod 9)

Step 4 — combine.

{11}^{1011}+{1011}^{11}\equiv 8+0=8 (\mod 9)

Answer: Option 2 — 8