Latest TCS Aptitude Question SOLUTION: f(x)= 2(x) +1, f(f(f(f….f(n))))12 times = 81919. Then n=?

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f(x)= 2(x) +1, f(f(f(f….f(n))))12 times = 81919. Then n=?

Asked In TCS Sanjoy (8 years ago)

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Deepak kumar singh  (8 years ago)

f(x)=2(x)+1
f(f(x))=2^2(x)+3
=2^2(x)+(2^2-1)
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f(f(f(f....(x)))) 12 times= 2^12(x)+(2^12-1)
given data from question
=> 819119=2^12(x)+2^12-1
=>819120=2^12(x+1)
=>20*2^12=2^12(x+1)
=>x=19 answer.

Is this solution Helpfull? Yes (28) | No (1)  

chanchal chauhan  (8 years ago)

f(x)=2(x)+1
f(n)=2n+1
f(f(n))=2(2n+1)+1=4n+3
f(f(f(n)))=2(4n+3)+1=8n+7
f(f(f(f(n))))=2(8n+7)+1=16n+15
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.
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f(f(f(f.....f(n))))12times= 4096n+4095
4096n+4095=81919
n=19

Is this solution Helpfull? Yes (12) | No   

sunil amritiya  (8 years ago)

ans. n=40959

firstly put x=19
f(x)=2x=1=39.........................1 time
f(x)=2*39+1=79.......................2 time
f(x)=2*79+1=159...................3time
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f(x)=2*40959+1=81919.........12 time

ans is n=40959

Is this solution Helpfull? Yes (6) | No (3)  

Om Prakash  (8 years ago)

Given, f(x)=2(x) + 1 = 2^1(x) +1
f(f(x))=f2(x)=2[(2(x)+1]+1= 4(x) + 2 +1 = 2^2(x) + 2 + 1
f(f(f(x)))=f3(x)=4[2(x) + 1] + 2+ 1= 8(x) + 4+ 2 +1 = 2^3(x) + 4 + 2 + 1
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f12(x)= 2^12(n) + ( 2^12 + 2^11 + 2^10 + ................ + 2^3 + 2^2 + 2^1 + 1) ---> in Gp
f12(x)=4096(n) + [ (2^12 - 1)/(2 - 1)]
since f12(x) =81919
hence,
4096(n) + 4096 -1=81919
after simplify...
n=19

Is this solution Helpfull? Yes (3) | No   

Karuna Nidhan  (8 years ago)

19,

simulate the function for three times and you will have an expression equate it with the value and the answer is out.

in this question 2^t*(n+1)=81920(i.e val+1)

Is this solution Helpfull? Yes (2) | No (1)  

Deepesh Malviya  (8 years ago)

First time: (81919-1)/2=40959
Second time: (40959-1)/2=20479
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.
Twelfth time: (39-1)/2=19
Thus answer is 19

Is this solution Helpfull? Yes (1) | No   

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