Latest Elitmus Aptitude Question SOLUTION: If for any series b(n)=2b(n-1) when b(n) is odd number and b(n)=b(n-1) if b(n) is even number. then find value of b(100)-b(97)-b(96) if b(0)=1

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(#M40151470) ELITMUS Numerical Ability question Keep an EYE

If for any series b(n)=2b(n-1) when b(n) is odd number and b(n)=b(n-1) if b(n) is even number. then find value of b(100)-b(97)-b(96) if b(0)=1

Asked In Elitmus ankit (9 years ago)

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Ramdas Singathiya  (9 years ago)

2^48.
b(100)=b(99)=2b(98)=2b(97)................(1)
so b(100)-b(97)-b(96)= b(97)-b(96) = 2b(96)-b(96)=b(96)

Now 96 is even so b(96)=b(95)....
b(95)=(2^1) * b(94)=(2^1) * b(93) =(2^2)* b(91)=(2^3)*b(89)

so here we can realize that 95-2(x) will have 2^x in multiply.... so finally 95-2x=1 => x=47

so finally it will be (2^47) * b(1)=(2^47) * (2)b(0) =2^48 !!!!!!!!!!!! :) I hope this is the answer in given options. right ankit?

Is this solution Helpfull? Yes (20) | No (4)  

Jayant Kumar Nayak  (8 years ago)

answer is 2^48

b(100)=b(99)=2^1(b98)=2^1b(97)=2^2b(96)=2^2b(95) ....
so we can write it as 2^(number of odd numbers starting from 1 to 100)
no. of odd numbers = 100/2 = 50
so, b(100)=2^50

simillarly ,
b(97) = 2^(97/2 + 1) =2^49
b(96) = 2^(96/2)=2^48

therefore b(100) - b(97) - b(96)
= 2^50 - 2^49 -2^48
= 2^48(2^2 - 2 - 1)
=2^48

Is this solution Helpfull? Yes (10) | No   

soni kumari  (9 years ago)

2^48 b(0)=2^48

Is this solution Helpfull? Yes (3) | No (2)  

MR. X  (9 years ago)

b(96)=2^48

Is this solution Helpfull? Yes (3) | No (2)  

satyadev  (9 years ago)

we can simple take it like b(100)=2^50 as per given equation and b(97)=2^49 and b(96)=b(95)=2^48
thn (2^50 - 2^49) = 2^49

nd now 2^49 -2^48= 2^48

so 2^48 is requires answer

Is this solution Helpfull? Yes (3) | No (1)  

sai venkata krishna  (8 years ago)

b(0)=1
b(1)=2(b(0))=2*1=2
b(2)=b(2-1)=b(1)=2
b(3)=2(b(3-1)=2(b(2))=2(2)=4
.
.
.
.
b(100)=b(99)
b(99)=2b(98)
b(98)=b(97)
b(97)=2b(96)
b(96)=b(95)
================b(100)-b(97)-b(96)==>4b(95)-2b(95)-b(95)==>b(95)
b(95)=2b(94)
b(95)==96
ANS:- 96

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

Ehtesham  (9 years ago)

b(100)-b(97)-b(96)
=b(99)-2b(96)-(95)
=2b(98)-2b(95)-2b(94)

since ,2 is eing multiplied whenver odd to even conversion takes place inside b().
so count no of even between 2 to 98 =49
so ,no of 2's will be 2^49 for b(100).

similarly ,for 2^48 for b(97),and 2^47 for b(96)

now solve it
2^49-2^48-2^47
=2^47(4-2-1)
=2^47

Is this solution Helpfull? Yes (1) | No (3)  

sangam jain  (8 years ago)

question is wrong since it should be like "n is even or n is odd not b(n))"

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