Browse · MATH Print → jmc algebra intermediate Problem Find 221⋅441⋅881⋅16161⋯. Solution — click to reveal We can write 221⋅441⋅881⋅16161⋯=221⋅22⋅41⋅23⋅81⋅24⋅161⋯=221+42+83+164+⋯.Let S=21+42+83+164+⋯.Then 2S=1+22+43+84+⋯.Subtracting these equations, we get S=1+21+41+81+⋯=1−1/21=2,so 221⋅441⋅881⋅16161⋯=2S=22=4. Final answer 4 ← Previous problem Next problem →