50ᵗʰ root of a number x is 12 and 50ᵗʰ root of another number y is 18 . Then the remainder obtained on dividing (x+y) by 25 is ____________.

Solution

Maths sir · 4 steps

Write x = 12⁵⁰ and y = 18⁵⁰, then hunt for a power that sits one below a multiple of 25.

1. Find x and y. The 50th root of x is 12, so x = 12⁵⁰. In the same way, y = 18⁵⁰. We need 12⁵⁰ + 18⁵⁰ modulo 25.
2. Handle y. 18² = 324 = 325 - 1, and 325 = 13 × 25. So 18⁵⁰ = (325 - 1)²⁵. In its binomial expansion every term except the last has a factor 325, so 18⁵⁰ = 25k + (-1)²⁵ = 25k - 1.
3. Handle x. 12⁵ = 248832 = 25 × 9953 + 7, so 12¹⁰ = (25 × 9953 + 7)², which leaves the same remainder as 7² = 49. Since 49 = 50 - 1, we can write 12¹⁰ = 25a - 1. Then 12⁵⁰ = (25a - 1)⁵, and the binomial expansion gives 25b - 1.

   x + y = (25b - 1) + (25k - 1) = 25(b + k) - 2 = 25(b + k - 1) + 23

4. So. A remainder must lie between 0 and 24, so -2 becomes 25 - 2. The remainder is 23.

The common slip: 2. Getting x + y = 25m - 2 and then writing the remainder as 2 gives 2. A remainder cannot be negative, so -2 must be turned into 25 - 2 = 23.

The idea to keep. For remainders of huge powers, look for a power of the base that is 1 more or 1 less than a multiple of the divisor. Expand with the binomial theorem and only the last term survives.

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