Questions in MBA entrance exams are based on two concepts:
1. Conversion of numbers from one base system to another.
2. Arithmetic Operations of numbers in any base system.
(a) (10.25)_{10}_{ }
_{}
Note: Keep multiplying the fractional part with 2 until decimal part 0.00 is obtained.
(0.25)_{10} = (0.01)_{2}
Answer: (10.25)_{10} = (1010.01)_{2}
(b) (28.3125)_{10}
The given number has 2 parts:
(i) Conversion of an integral part:
(28)_{10} = (11100)_{2}
(ii) Conversion of the fractional part:
Note: We should stop multiplying the factorial part by 2, once we get 0 as a fraction or the fractional part is nonterminating. It can be decided depending on the number of digits in the fractional part required.
(a) (4324.235)_{10}
(i) Integral Part
(4324)_{10 }= (10344)_{8}
(ii) Fractional Part
Here, we need to express every digit of octal number into its binary form comprising of 3 digits.
Note: Using the logic discussed in points 7 to 10, we can do direct conversions in any two bases a and b such that a=b^{n} where we will form blocks of n digits when the number is in base b and then write its decimal equivalent.
Example: For conversion from base 3 to base 9, we need to make blocks of 2 digits as 9 = 3^{2}, for instance (22112)_{3} = (02 21 12)_{3} = (275)_{9}
Let us start with an easy example to understand the rationale.
(a) (8358)_{10} + (5684)_{10}
(8358)_{10} + (5684)_{10} = (14042)_{10}
➢ Logic
(b) (3542)_{6} + (4124)_{6}
(3542)_{6 }+ (4124)_{6} = (12110)_{6}
(a) (237)_{10} – (199)_{10}
(237)_{10 }– (199)_{10} = (38)_{10}
➢ Logic
(b) (422)_{5} – (243)_{5}
(422)_{5} – (243)_{5} = (124)_{5}
For multiplying numbers in any base system, multiply them as we normally do for decimal numbers and while writing, write each number in the given base system
Example: Calculate (52)8 × (6)8
Solution:
Given, (52)_{8} × (6)_{8}
We first multiply 6 with 2 i.e., 2 × 6 = 12 and write it in base 8 = (14)_{8.}
Now, 4 will be the unit's digit of the final answer and 1 will be carried forward.
Now we multiply 6 with 5 and add any carry forward i.e., 5 × 6 + 1 = 31 and write it in base 8 = (37)_{6}
Now, 37 will be the leftmost digit of the final answer.
∴ (52)_{8} × (6)_{8} = (374)_{8}
(a) Is (7364)_{9 }divisible by 8?
7 + 3 + 6 + 4 = 20 which is not divisible by 8. Hence, the given number is not divisible by 8.
Rule: (x)_{b} is divisible by (b1) if all the digits of (x)_{b} add up to be divisible by (b1).
(b) Is (5236)_{9} divisible by 10?
Rule: (x)_{b} is divisible by (b+1) if the difference of the sums of alternate digits of (x)_{b} is either 0 or divisible by (b+1).
(c) What is the IGP (Index of Greatest Power) of 9 in (780)_{9}?
Rule: For a number in base b, if there are k zeroes in the end then it is divisible by b^{k}. Also, k is the IGP of b in the number.
 (15AA51)_{19} = 1*19^{5}+ 5*19^{4}+ 10*19^{3}+ 10*19^{2}+ 5*19^{1}+ 1*19^{0 }= (19+1)^{5} = 20^{5 }(Using binomial theorem)
 Therefore, the fifth root is 20.
Other examples of similar kind are:
Example 2. How many 4digit numbers in base 9 are perfect squares?
 First, we need to know the range of 4digit numbers in base 9
Least 4 digit number possible= (1000)_{9 }= 9^{3} =729
 Observation: Lowest n digit number in base k = k^{(n1)}
Highest 4 digit number possible= (8888)_{9} = 9^{4}1= 6560
 Observation: Highest n digit number in base k = k^{n}1
 From 729 to 6560, the squares vary from 27^{2} to 80^{2}.
 Number of perfect squares present = 80  26 = 54.
Correct Answer is Option (a).
 (ab)^{2} = ccb, the greatest possible value of ‘ab’ can be 31, since 31^{2} = 961 (and since ccb > 300), 300 < ccb < 961, so 18 < ab < 31.
 So the possible value of ab which satisfies (ab)^{2} = ccb is 21.
 So 21^{2} = 441, ∴ a = 2, b = 1, c = 4.
Q.2. Convert the number 1982 from base 10 to base 12. The result is? (CAT 2000)
(a) 1182
(b) 1912
(c) 1192
(d) 1292
Correct Answer is Option (c).
Q.3. In a number system, the product of 44 and 11 is 1034. The number 3111 of this system, when converted to the decimal number system, becomes? (XAT 2001)
(a) 406
(b) 1086
(c) 213
(d) 691
(e) None of the above
Correct Answer is Option (a).
 Let the base be n
 (4n+4)(n+1) = n^{3}+3n+4
 n^{3}4n^{2}5n = 0
 n(n5)(n+1) = 0
 n = 5
 (3111)_{5 }= (406)_{10}
Q.4.The product of two numbers 231 and ABA is BA4AA in a certain base system (where the base is less than 10), where A and B are distinct digits. What is the base of that system? (CAT 2010)
(a) 5
(b) 6
(c) 7
(d) 8
(e) 4
Correct Answer is Option (b).
Hence ,we can write (b+4+a)2a = 0 or (base)+ 1 (let’s take (base+1))
i.e. b = base + a – 3 231 * aba = 2a(base)^{4}+(2b+3a)(base)^{3}+(3a+3b)(base)^{2}+(3a+b)(base)+a
 Now put b = base + a – 3 , in above equation and Compare it with ba4aa, We get:
 2a + 2 = b
 4a – 3 = a
 Solving them gives a = 1 , b = 4
 Hence, base = b  a + 3 = 6
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1. What is a base system? 
2. How do you convert a number from one base system to another? 
3. How do you perform arithmetic operations in a specific base system? 
4. What are some important concepts of the base system? 
5. Can you provide some previous year questions based on the base system? 
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