Oracle
Company
Numerical Ability
Age Problem
if A=2^65 AND B=2^64+2^63.....+2^0
options
a) A is larger than B by 2^64 times
b)B is larger than A by 1
c)A is larger than B by 1
d) None
Read Solution (Total 12)
-
- Ans) C
lets say A : 2^4=16 , B: 2^3+2^2+2^1+2^0 = 15 so A is larger than B by 1 always - 9 years agoHelpfull: Yes(14) No(1)
- ans) c
B=2^64+2^63+........+2^0 (here makes a GP series)
so, sum of GP series=a(r^n-1)/(r-1)
=1(2^64-1)/(2-1)
=2^64-1
and, A=2^64 (which is larger than by 1)
- 9 years agoHelpfull: Yes(4) No(2)
- multiply B by 2 on both sides by 2 then do A-B u will get A-B=1
- 9 years agoHelpfull: Yes(3) No(0)
- B is a GP with
a = 2^0 = 1 and r = 2 and n = 65
therefore B = 2^0 (2^65 - 1) / (2 -1) = 2^65 - 1 = A - 1
thus ans is C) A is larger than B by 1 - 9 years agoHelpfull: Yes(2) No(0)
- option c a is larger than b by 1
- 9 years agoHelpfull: Yes(1) No(0)
- take them as binary no. A=100000000...00(64 0's) and 011111..11(64 1's) so A>B by 1.
eg, 100 in binary =4 in decimal
011 in binary =3 in decimal
the same rule applies over here.
so ,option c. - 7 years agoHelpfull: Yes(1) No(0)
- Ans: A is Larger than B by 1
- 9 years agoHelpfull: Yes(0) No(0)
- Answer is C
Since, if we take a small no say
A= 2^3 and B = 2^2+2^1 + 2^0
then is is clear that "A is larger than B by 1" - 9 years agoHelpfull: Yes(0) No(0)
- c)A is larger than B by 1
- 9 years agoHelpfull: Yes(0) No(0)
- 2^n-2^n-1=1
Hence a is larger than b by 1. - 9 years agoHelpfull: Yes(0) No(1)
- A=2^65
B=2^65-1(after solve GP)
so that A is larger than B by 1. - 9 years agoHelpfull: Yes(0) No(0)
- c option
coz B= 2^65-1 by gp series summtion - 8 years agoHelpfull: Yes(0) No(0)
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