The remainder on dividing 5⁹⁹ by 11 is ____________.

Solution

Maths sir · 4 steps

Find a power of 5 that is one more than a multiple of 11, then let the binomial expansion do the rest.

1. Spot the useful power. Try small powers of 5 until one lands just above a multiple of 11. 5⁵ = 3125 = 3124 + 1, and 3124 = 11 × 284.
2. Split the exponent. Since 99 = 5 × 19 + 4, write 5⁹⁹ = 5⁴ × (5⁵)¹⁹ = 625 × (1 + 3124)¹⁹.

   (1 + 3124)¹⁹ = 1 + C(19, 1)3124 + C(19, 2)3124² + … = 1 + 11k

3. Use the expansion. Every term after the first contains 3124, which is a multiple of 11. So 5⁹⁹ = 625(1 + 11k) = 625 + 11 × 625k, and the remainder comes only from 625.
4. So. 625 = 11 × 56 + 9, so the remainder on dividing 5⁹⁹ by 11 is 9.

The common slip: 1. Forgetting the leftover factor 5⁴ and keeping only (5⁵)¹⁹ gives 1. The 625 in front still has to be divided by 11.

The idea to keep. For a remainder, write the base power as (multiple of the divisor + 1). In the binomial expansion every term except the 1 is divisible, so only the leftover factor decides the remainder.

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