The remainder, when 19²⁰⁰+23²⁰⁰ is divided by 49 , is ___________.

Solution

Maths sir · 4 steps

Write 19 and 23 as 21 minus 2 and 21 plus 2, and the odd terms cancel each other.

1. Pair them up. 19 = 21 - 2 and 23 = 21 + 2. When you add (21-2)²⁰⁰ + (21+2)²⁰⁰, every term with an odd power of 21 cancels, and every even-power term appears twice.

   19²⁰⁰+23²⁰⁰ = 2[2²⁰⁰ + C(200, 2)21²· 2¹⁹⁸ + C(200, 4)21⁴· 2¹⁹⁶ + …]

2. Drop multiples of 49. Every term after the first has a factor 21² = 441 = 9 × 49, so it leaves remainder 0. That leaves 19²⁰⁰+23²⁰⁰ ≡ 2 × 2²⁰⁰ = 2²⁰¹ (mod 49).
3. Expand once more. 2²⁰¹ = 8⁶⁷ = (7+1)⁶⁷ = 1 + 67 × 7 + C(67, 2)7² + …. Every term from 7² onward is a multiple of 49, so 2²⁰¹ ≡ 1 + 469 (mod 49).
4. So. 470 = 9 × 49 + 29, so the remainder is 29.

The common slip: 39. Forgetting the factor 2 in front of the bracket leaves 2²⁰⁰, which gives 39 instead of 29.

The idea to keep. For a remainder on division by m, write each base as a multiple of a factor of m plus or minus a small number. Expand, and every term carrying enough of that factor drops out.

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