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JEE Main
Exam
JEE Main
Subject
Mathematics
Chapter
Basics of Mathematics

JEE Main · Mathematics · Basics of Mathematics

If the greatest divisor of 30 · 5^ 2 n +4 · 2^ 3 n is p, n ∈ N and...

If the greatest divisor of 30⋅52n+4⋅23n30 \cdot 5^{2 n}+4 \cdot 2^{3 n} is p,∀n∈Np, \forall n \in N and that of 22n+1−6n−22^{2 n+1}-6 n-2 is q,∀n∈Nq, \forall n \in N, then p+q=p+q=

Options

  1. A26
  2. B52
  3. C104
  4. D13

Correct answer

B. 52

Step-by-step solution

We have, 30⋅52n+4⋅23n30 \cdot 5^{2 n}+4 \cdot 2^{3 n} Put  Put \text { Put }n=130×25+4×8=750+32=782782=2×23×17\begin{aligned}n & =1 \\ 30 \times25+4 \times8 & =750+32=782 \\ 782 & =2 \times23 \times17 \end{aligned}PutPutx=230 \times 625+4 \times 64=18750+256=19006InEqs.(i)and(ii),wegetGreatestdivisorIn Eqs. (i) and (ii), we get Greatest divisor=2 \times 17=34∴p=34Nowin\therefore \quad p=34 Now in2^{2 n+1}-6 n-2PutPutn=18−6−2=0Putn=232−12−2=18Put8-6-2=0 Put n=2 32-12-2=18 Putn=3128−18−2=108128-18-2=108\becauseGreatestdivisorGreatest divisor=18∴\quadq=18∴p+q=34+18=52\begin{aligned} \therefore \quadq & =18 \\ \therefore p+q=34+18 & =52 \end{aligned}
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