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Understanding Base System Concepts and Questions for CAT Exam

Tuesday, September 22nd, 2020

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UNDERSTANDING BASE SYSTEMS:

The numbers that we commonly use are the decimal number. The system is called the decimal system. Why is it called the decimal system ? It is because it has 10 symbols – 0 , 1 , 2 , 3 , 4 , 5 ,6 , 7 , 8 , 9 .

For example : 85 means – 8 * 10^1 + 5 * 10^0

Here , 10 is known as the base of the decimal system.

In this article , we will learn about other bases , how to express various numbers in such bases and questions according to CAT , XAT and OMETs.

BASE :  It is the number of distinct symbols used in a particular number system.

For example

So, depending on the number of digits in the base system, there are many other systems possible. Have a look at the following table :

Number system	Base	Symbol
Binary	2	0,1
Septenuary	7	0,1,2,3,4,5,6
Octal	8	0,1,2,3,4,5,6,7
Decimal	10	0,1,2,3,4,5,6,7,8,9
Duo-decimal	12	0,1,2,3,4,5,6,7,8,9,A,B
Hexa-decimal	16	0,1,2,3,4,5,6,7,8,9,A,B,C,D,E,F

n the questions involving base systems , we are expected to figure out the different number of digits in that system of counting.

For example : If we had 7 digits instead of 10 ie. 0 ,1 ,2 ,3 ,4 ,5 , 6 , we will have system of counting as follows : 0 , 1, 2 ,3 , 4  5 ,6 , 10 , 11 , 12 ,13 , 14… .

In above system , 45 will be expanded as = 4 * 91 +  5 * 90

Conversions from one Base System to Another

1. Decimal to Binary

a.) (172)₁₀

b.) (28.3125)₁₀

The given number has 2 parts

1. Conversion of integral part

(28)₁₀ = (11100)₂

ii.) Conversion of the fractional part:

iii.) Multiply the decimal part with 2 successively and take the integral part of all the products starting from the first.

iv.) 0.3125*2=0.6250 -> (0)

v.) 0.6250*2=1.2500 -> (1)

0.2500*2=0.500 -> (0)

0.500*2=1 -> (1)

Therefore, (0.3125)₁₀ = (0.0101)₂

(28.3125)₁₀ = (11100.0101)₂

NOTE: We should stop multiplying the factorial part by 2, once we get 0 as a fraction or the fractional part is non-terminating. It can be decided depending on the number of digits in the fractional part required.

2. Binary to Decimal

1. (110101)₂

=1*2⁵+ 1*2⁴+ 0*2³+ 1*2²+ 0*2¹+ 1*2⁰ = (53)₁₀

2. (0.11001)₂

=1*2^-1+ 1*2^-2+ 0*2^-3+ 0*2^-4+ 1*2^-5 = (0.78125)₁₀

3. Decimal to Octal

1. (4324.235)₁₀

1. Integral part

(4324)₁₀ = (10344)₈

ii.) Fractional part

iii.) 0.235*8=1.88 -> (1)

iv.)  0.88*8=7.04 -> (7)

v.) 0.04*8=0.32 -> (0)

We can stop here as the fraction is non-terminating.

(4324.235)₁₀ = (10344.170)₈

4. Octal to Decimal

1. (677.47)₈

= 6*8²+ 7*8¹+ 7*8⁰+ 4*8^-1+ 7*8^-2 = (447.609375)₁₀

5. Decimal to Hexadecimal

1. (23524)₁₀

(23524)₁₀ = (5BE4)₁₆

6. Hexadecimal to Decimal

1. (62A)₁₆

= 6*16²+ 2*16¹+ 10*16⁰ = (1578)₁₀

Now, we will look at conversions which do not include base 10. We can observe that following cases include 2 bases (let them be a and b), where a=bⁿ.

7. Binary to Octal

8 (octal) is cube of 2 (binary). So, to convert binary to octal we just need to club three digits of binary number starting from unit digit and write the decimal equivalent of each group.

1. (1101001011)₂ = (001 101 001 011)₂

(We can introduce two zeroes to form groups of 3 without changing the magnitude of the number)

(001 101 001 011)₂ = (1513)₈

8. Octal to Binary

Here, we need to express every digit of octal number into its binary form comprising of 3 digits.

1. (645)₈

6 = (110)₂

4 = (100)₂

5 = (101)₂

(645)₈ = (110 100 101)₂

9. Binary to Hexa-decimal

It is similar to the method discussed in 7^th point. Instead of clubbing 3, we club 4 digits.

1. (10101110)₂ = (1010 1110)₂ = (AE)₁₆

10. Hexa-decimal to Binary

We need to express every digit of Hexa-decimal into its binary form comprising of 4 digits.

1. (3A91)₁₆

3 = (0011)₂

A = (1010)₂

9 = (1001)₂

1 = (0001)₂

(3A91)₁₆ = (11101010010001)₂

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=bⁿ where we will form blocks of n digits when the number is in base b and then write its decimal equivalent. Ex- for conversion from base 3 to base 9, we need to make blocks of 2 digits as 9=3², for instance- (22112)₃ = (02 21 12)₃ = (275)₉

Arithmetic Operations while operating in a specific Base System

One method of performing arithmetic operations is to first convert the numbers to decimal system, perform the required operations and then convert the numbers back to required base.

However arithmetic operations can also be directly done given that numbers are expressed in the same base.

Addition –

Let us start with an easy example to understand the rationale.

1. (8358)₁₀ + (5684)₁₀

Carry over	1 (quotient when 10 is divided by 10)	1 (quotient when 14 is divided by 10)	1 (quotient when 12 is divided by 10)
	8	3	5	8
+	5	6	8	4
	(14)	(10)	(14)	(12)
1 (quotient when 14 is divided by 10)	4 (Remainder when 14 is divided by 10)	0 (Remainder when 10 is divided by 10)	4 (Remainder when 14 is divided by 10)	2 (Remainder when 12 is divided by 10)

(8358)₁₀ + (5684)₁₀ = (14042)₁₀

 

LOGIC –

1. Start from unit position; 8+4=12, and 12 when divided by 10 (base) gives us remainder as 2, which is the total for that column. 12 when divided by 10 gives us quotient as 1 which is carried over to next column.
2. In second column, 1+5+8=14. When 14 is divided by 10 (base), we get 4 as remainder (total for that column) and 1 as quotient which is carried over to next column.
3. In third column, 1+3+6=10. 10 when divided by 10 gives us 0 as remainder and 1 as quotient. We proceed in similar and get the required sum.

2. (3542)₆ + (4124)₆

Carry over	1 (quotient when 7 is divided by 6)	1 (quotient when 7 is divided by 6)	1 (quotient when 6 is divided by 6)
	3	5	4	2
+	4	1	2	4
	(8)	(7)	(7)	(6)
1 (quotient when 7 is divided by 6)	2 (Remainder when 8 is divided by 6)	1 (Remainder when 7 is divided by 6)	1 (Remainder when 7 is divided by 6)	0 (Remainder when 6 is divided by 6)

(3542)₆ + (4124)₆ = (12110)₆

 

SUBTRACTION –

1. (237)₁₀ – (199)₁₀

 

	-1 (borrowed to tens place)	-1+10 (1 borrowed to unit place and 10 borrowed from hundreds place)	+10 (borrow from tens place)
	2	3	7
–	1	9	9
	(2-1 – 1=0)	(3-1+10 – 9=3)	(7+10 – 9= 8)
	0	3	8

(237)₁₀ – (199)₁₀ = (38)₁₀

 

Logic –

1. In units place 7<9, hence we borrow 10 (base) from tens place. 17-9=8
2. Tens place reduces from 3 to 2. Still 2<9, hence we borrow 10 (base) from hundreds place. 12-9=3
3. Hundreds place reduces to 1 (as we borrowed from it in earlier step). 1-1=0

 

2. (422)₅ – (243)₅

 

	-1 (borrowed to tens place)	-1+5 (1 borrowed to units place and 5 borrowed from hundreds place)	+5 (borrow from tens place)
	4	2	2
–	2	4	3
	(4-1 – 2=1)	(2-1+5 – 4=2)	(2+5 – 3= 4)
	1	2	4

(422)₅ – (243)₅ = (124)₅

MULTIPLICATION –

Using the analogy of base 10, we can multiply in other bases too

1. (346)₇ * (4)₇

Carry over	2 (quotient when 19 is divided by 7)	3 (quotient when 24 is divided by 7)
	3	4	6
+			4
	(14)	(19)	(24)
2 (quotient when 14 is divided by 7)	0 (Remainder when 14 is divided by 7)	5 (Remainder when 19 is divided by 7)	3 (Remainder when 24 is divided by 7)

(346)₇ * (4)₇ = (2053)₇

 

2. (76)₈ * (45)₈

			7	6
*			4	5
		4	6	6	Row obtained by (76)8*(5)8
+	3	7	0	 x	Row obtained by (76)8*(4)8
	4	3	6	6	Adding (466)8 and (3700)8

76)₈ * (45)₈ = (4366)₈

DIVISIBILITY –

1) Is (7364)₉ divisible by 8?

Solution- The logic behind this question is same as checking the divisibility of any number (in decimal system) by 9. We add the digits and then check the divisibility.

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 (b-1) if all the digits of (x)_b add up to be divisible by (b-1).

2) Is (5236)₉ divisible by 10?

Solution- The logic behind this question is same as checking the divisibility of any number (in decimal system) by 11. We first find the sum of alternate digits and then find the difference of the sums obtained. This difference should either be divisible by 0 or divisible by 11(or 10 in the case of this question).

5+3=8; 2+6=8

8-8= 0. Hence, the number is 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).

3) What is the IGP (Index of Greatest Power) of 9 in (780)₉ ?

Solution– (780)₉ = 7*9²+ 8*9¹+ 0*9⁰ = 9(7*9+ 8)

Thus IGP= 1. As the highest power of 9 with which the number is divisible is 1

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.

Important Concepts of Base System

1) Find the fifth root of (15AA51)₁₉

Solution- (15AA51)₁₉ = 1*19⁵+ 5*19⁴+ 10*19³+ 10*19²+ 5*19¹+ 1*19⁰

= (19+1)⁵ = 20⁵ (Using binomial theorem)

Therefore, the fifth root is 20

Other examples of similar kind-

(121)_n = n²+ 2n+ 1= (n+1)² (n>2)

(1331)_n = n³+ 3n²+ 3n+ 1= (n+1)³ (n>3)

(14641)_n, (15AA51)_n and so on (the digits of the numbers used in the above examples form Pascal’s triangle)…

2) How many 4-digit numbers in base 9 are perfect squares?

Solution- First we need to know the range of 4-digit numbers in base 9

Least 4 digit number possible= (1000)₉ = 9³ =729

OBSERVATION- Lowest n digit number in base k= k^(n-1)

Highest 4 digit number possible= (8888)₉ = 9⁴-1= 6560

OBSERVATION- Highest n digit number in base k= kⁿ-1

From 729 to 6560, the squares vary from 27² to 80² .

Number of perfect squares present= 80-26=54.

Previous Year Questions based on Base System

1) Let a, b, c be distinct digits. Consider a two-digit number ‘ab’ and a three-digit number ‘ccb’, both defined under the usual decimal number system, if (ab)² = ccb > 300, then the value of b is? (CAT 1999)

a. 1

b. 0

c. 5

d. 6

Solution– b=1

(ab)² = ccb, the greatest possible value of ‘ab’ can be 31, since 31² = 961 (and since ccb > 300), 300 < ccb < 961, so 18 < ab < 31.

So the possible value of ab which statisfies (ab)² = ccb is 21. So 21² = 441, ∴ a = 2, b = 1, c = 4.

2) Convert the number 1982 from base 10 to base 12. The result is? (CAT 2000)

(1) 1182

(2) 1912

(3) 1192

(4) 1292

Solution- (3)-1192

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)

(1) 406

(2) 1086

(3) 213

(4) 691

(5) None of the above

Solution- (1) 406

Let the base be n

(4n+4)(n+1)= n³+3n+4

=> n³-4n²-5n=0

=> n(n-5)(n+1)=0

n=5

(3111)₅ = (406)₁₀

4)Two numbers in the base system B are 2061 and 601 . The sum of these 2 numbers in decimal system is 432. Find the value of 1010 in decimal system:

a)110

b)120

c)130

d)140

e)150

Solution : Option C is the correct answer .

NOTE : THIS QUESTION IS TECHNICALLY WRONG BECAUSE WE CAN NOT HAVE 6 IN BASE 5.

Converting numbers in decimal system :

2061 = 2B^3 + 6B + 1

601 = 6B^2 + 1

solving the above equation :

B=5.

Therefore , 1010 gives 130 in decimal system.

5)The product of two numbers 231 and ABA is BA4AA in a certain base system (where 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

Solution : 231 is a multiple of 11

Hence ,we can write (b+4+a)-2a = 0 or (base)+ 1 ( let’s take (base+1) )

ie. 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

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

All The Best!

You can also see Reading Comprehension for CAT – Different types of Questions

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