If a number x is in octal form and having zero as unit place then if we change that number x in decimal form then what is the probability that number x having zero as unit place?
Options.
A) 11/64
B) 11/69
C) 11/66
D) 11/64
E) 11/61
example (8)10=(10)8(8)10=(10)8, (16)10=(20)8(16)10=(20)8, (24)10=(30)8(24)10=(30)8, (32)10=(40)8(32)10=(40)8, (40)10=(50)8(40)10=(50)8, . . .   So out of five numbers end with zero in octal form, only one (50)8=(40)10(50)8=(40)10 has zero in decimal format. So the required probability is 1/5   My initial thinking:   Let the number be "ab0" in octal form. Remember here a, b &lt; 8. So total possibilities are 64.   If we convert the given number into decimal format, then (ab0)8 = 64a + 8b.   We will find for how many values of 64a + 8b ends in zero.   64a + 8b = 60a + (4a + 8b)   For (0,0), (1, 2), (1, 7), (2, 4), (3, 1), (3, 6), (4, 3), (5, 5), (6, 2), (6, 7), (7, 4), (0, 0) the above expression gives unit digit 0.   So required probability = 11/64   (Imp: Still I am not comfortable with the solution I have written as it is not specified whether the given number is 3 digit or not. Just learn how I tried to solve it. )   11/64nswer:11/64
<result>A</result>
If a number x is in octal form and having zero as unit place then if we change that number x in decimal form then what is the probability that number x having zero as unit place?