The remainder on dividing 1 + 3 + 3² + 3³ + ..... + 3²⁰²¹ by 50 is _________.

Solution

Maths sir · 4 steps

Sum the GP first, then write 3²⁰²² as (10-1)¹⁰¹¹ and keep only the last two digits.

1. Sum the GP. The terms run from 3⁰ to 3²⁰²¹, so there are 2022 terms with ratio 3. The sum is S = (3²⁰²²-1)/(3-1).

   S = (3²⁰²²-1)/2 = (9¹⁰¹¹-1)/2 = ((10-1)¹⁰¹¹-1)/2

2. Binomial expansion. In (10-1)¹⁰¹¹, every term with 10² or a higher power is a multiple of 100. Since 1011 is odd, the last two terms are C(1011, 1) × 10 - 1 = 10110 - 1 = 10109. So 9¹⁰¹¹ = 100m + 10109 for some integer m.
3. Halve it. S = (100m + 10108)/2 = 50m + 5054. Here 5054 = 50 × 101 + 4, so S = 50(m + 101) + 4.
4. So. The remainder on dividing the sum by 50 is 4.

The common slip: 1. Taking 2021 terms and writing the sum as (3²⁰²¹-1)/2 gives 1, not 4. The series starts at 3⁰, so it has 2022 terms.

The idea to keep. For a remainder of N/2 by 50, find N modulo 100 first. Write the base as a number near a power of 10, like 9 = 10-1, and only the last two binomial terms matter.

Answer from the NTA final answer key. Solution by PrepWiser.