Quantrex Quantrex AcademyJEE · NEET · NDA PYQs with solutions Open app
JEE MainMathematicsBinomial Theorem

If the fractional part of the number 3²⁰² 17 is k 17 , where k is an integer such that 0 k < 17 , then the value of k is :

Options

  1. A9
  2. B16
  3. C8
  4. D1

Correct answer

C. 8

Step-by-step solution

The fractional part of 3²⁰² 17 is given by k 17 , which means k is the remainder when 3²⁰² is divided by 17 . We can write 3²⁰² as 3^2 3²⁰⁰ = 9 (3^4)⁵⁰ . 3²⁰² = 9 (81)⁵⁰ Now, express 81 in terms of a multiple of 17 : 81 = 17 5 - 4 = 85 - 4 Using the binomial theorem, we have: 9 (85 - 4)⁵⁰ = 9 ( 17M + (-4)⁵⁰ ) Thus, the remainder depends on 9 (-4)⁵⁰ . 9 (-4)⁵⁰ = 9 4⁵⁰ = 9 (4^2)²⁵ = 9 (16)²⁵ Now, express 16 as (17 - 1) : 9 (17 - 1)²⁵ = 9 ( 17N + (-1)²⁵ ) 9 (17N - 1) = 17(9N) - 9 The remainder is -9 . Since the remain

Practice Binomial Theorem on Quantrex Academy →

More from Binomial Theorem

If 26 ( 2^3 3 12 2 + 2^5 5 12 4 + 2^7 7 12 6 + + 2¹³ 13 12 12 ) = 3¹³ - , then is equal to: 2026If (1 - x^3)¹⁰ = _ r=0 ¹⁰ a_r x^r (1-x)^ 30-2r , then 9a₉ a₁₀ is equal to __________. 2026If the coefficients of the middle terms in the binomial expansions of (1 + x)²⁶ and (1 - x)²⁸ , 0 , are equal, then the value of is: 2026The coefficient of x^2 in the expansion of (2x^2 + 1 x )¹⁰ , x 0 , is : 2026If the sum of the coefficients of x^7 and x¹⁴ in the expansion of ( 1 x^3 - x^4 )^n , x 0 , is zero, then the value of n is __________. 2026In the expansion of (9x- 1 3 x )¹⁸ , x>0 , if the term independent of x is (221)k , then k is equal to: 2026Let the smallest value of k N , for which the coefficient of x^3 in (1+x)^3 + (1+x)^4 + (1+x)^5 + + (1+x)⁹⁹ + (1+kx)¹⁰⁰ , x 0 , is (43n + 101 4 ) (¹⁰⁰C₃ ) for some n N , be p . The 2026If for 3 r 30 , 30 30-r + 3 30 31-r + 3 30 32-r + 30 33-r = m r , then m equals: 2026 Full Binomial Theorem list All JEE Main PYQs