We humans use decimal system for our mathematical calculations. We know that the computers work on binary system. Some of us who are from engineering or science background are also aware of octal and hexadecimal systems. But the base is not just limited to only these few; any number can be the base. It all depends upon the number of different digits we use in that system.
In decimal system the base is 10, and we use 0 – 9 digits. Similarly in base 2 we use 0 and 1 as digits and in hexadecimal system since the base is 16, we use 0-9 numbers and A – F alphabets (A, B, C, D, E and F represent values from 10 to 15 respectively) to have 16 different digits. Similarly, for new base say 5 we use the 0-4 digits and so on.
Note: If the number say 3241 is in base x, to avoid confusion it is represented by (3241)x.
The problems in competitive exams from base systems are based on two concepts:
- Conversion of a number from one base to another base
- Simple mathematical operations like addition, subtraction and multiplication of numbers in base other than 10.
Process of converting a number from base 10 into another base
When we convert a number from base 10 to any other base say x, we divide the number by x and whatever is the remainder becomes the unit digit of the number in base x and the quotient obtained is further divided and the remainder thus obtained becomes the ten’s digit in base x and so on. The process is repeated till the quotient becomes ZERO.
Example: Convert 435 into base 6.
Solution:
Dividing 435 by 6 the quotient is 72 and the remainder is 3. This remainder 3, is the unit digit of the number in base 6.
Again dividing the quotient 72 by 6, the new quotient of 12 and the remainder is 0. This remainder 0 is the ten’s place digit of the number in base 6. The graphical representation is given below for better understanding
Hence, the number 435 in base 10 = 2003 in base 6.
Or \({\left( {435} \right)_{10}} = {\left( {2003} \right)_6}\)
Process of converting a number from another base into base 10
The process is reverse of what we did in the above example. Instead of successive division, we do successive multiplication i.e.
Converting number \({\left( {abcd} \right)_p}\) into base 10 = \(d \times {p^0} + c \times {p^1} + b \times {p^2} + a \times {p^3}\)
Example: Convert \({\left( {3564} \right)_7}\) into base 10.
Solution:
\({\left( {3564} \right)_7}\)
\( = {\left( {3 \times {7^3} + 5 \times {7^2} + 6 \times {7^1} + 4 \times {7^0}} \right)_{10}}\)
\( = {\left( {1230} \right)_{10}}\)
Process of converting a decimal / Fraction into another base (say x)
Step 1: Multiply the decimal with the base x
Step 2: Remove the whole number part after the multiplication. This is the first digit after decimal in base x.
Step 3: Repeat step 2 till all the numbers after decimal get exhausted. The whole number after 2nd multiplication is the ten’s place digit, and after third multiplication, the whole number part is hundredth digits and so on.
Let us understand the working of above steps with the help of an example:
Example: Convert 0.256 from base 10 to base 5
Solution:
Following the above steps we get:
\(0.256 \times 5 = 1.28\) , The whole number part is 1.
\(0.28 \times 5 = 1.4\), The whole number part is 1.
\(0.4 \times 5 = 2.0\), The whole number part is 2.
Hence, \({\left( {0.256} \right)_{10}} = {\left( {0.112} \right)_5}\)
Process of converting a fraction from another base (say x) back to decimal
Say, 0.abc is in base x. The process of converting it into based 10 is as follow:
\({\left( {0.abc} \right)_x} = {\left( {a \times \frac{1}{x} + b \times \frac{1}{{{x^2}}} + c \times \frac{1}{{{x^3}}}} \right)_{10}}\)
For example, let us convert \({\left( {0.112} \right)_5}\) back to decimal.
\({\left( {0.112} \right)_5} = {\left( {1 \times \frac{1}{5} + 1 \times \frac{1}{{{5^2}}} + 2 \times \frac{1}{{{5^3}}}} \right)_{10}}\)
\( = {\left( {0.2 + 0.04 + 0.016} \right)_{10}}\)
\( = {\left( {0.256} \right)_{10}}\)
Addition, Subtraction and Multiplication in different bases
The process of addition, subtraction and multiplication in bases other than 10 might sound difficult because we haven’t done much practice of it during our school days. But the process is exactly same as we do in decimal system.
Let us see the process of addition of two numbers in decimal system, and then we will extend the logic to add two numbers in bases other than 10.
Say, we have to add 368 to 437 i.e
We start from right and add the unit digits, i.e. 8 + 7 = 15. Then we write 5 at the unit digit of the final answer and carry over the number 1 to next column of ten’s digits. We repeat the same process till the end.
The entire logic lies on the answer of the question that why did we carry over 1 and kept 5 at the unit place? The logic is, 15 can be written as 10+5 or \(1 \times 10 + 5\) (read 1 time 10 plus 5). So we kept 5 at the unit place and carried over 1.
(If the sum was say 47, it would have been written as \(4 \times 10 + 7\), so we would have kept 7 at unit place and carried over the number 4).
Hence the logic is, whenever the sum is greater than the base, we divide the sum by the base, keep the remainder and carry over the quotient.
Example: Calculate (456)7 + (234)7
Solution:
The addition of unit digits (6 and 4) is 10, which is more than 7, and would be written as \(1 \times 7 + 3\). The quotient is 1 and the remainder of 3. The Remainder is kept at the units place of the answer and the quotient gets carried over to the ten’s place. The process is repeated till the end.
Similarly, the process of subtraction and multiplication is same as explained above. Let us take few worked out examples.
Example: Calculate \({\left( {54} \right)_8} \times {\left( {36} \right)_8}\)
Solution:
First we multiple 6 with 4 = 24
24 = 3\( \times \)8 + 0, so the remainder 0 will be written at the unit place of the answer and 3 will be carried over to next column (ten’s digits column)
Next, we multiply 6 with 5 and add 3 (carry) = 33. Again, 33 = 4\( \times \)8 + 1 = (41)8. The Process till now looks like:
Similarly, 3 multiplied by 4 = 12 = 1\( \times \)8 + 4, so 4 is written in units place and 1 is carried over. Finally 3 multiplied by 5 + 1 (carry) = 16 = 2\( \times \)8 +0 = (20)8. The process till now looks like:
The last step is to add the two rows in base 8 to get the final answer which is equal to (2450)8
Example: Calculate (753)9 – (476)9
Solution:
The unit digits of the given numbers are 3 and 6, as 3 < 6 (we can’t subtract 6 from 3), we will take 1 carry from ten’s digit of the first number i.e 5 and the new ten’s place digit of the first number will become 4. As we are working in base 9, the weight-age of 1 carry is 9. So 3 + 1 (carry) = 3 + 9 = 12.
Now, 12 – 6 = 6 is the unit digit of the answer.
Similarly moving to the next column, the new ten’s digit of first number is 4 and ten’s digit of second number is 7, again we will repeat the same concept of carry as above. The final answer is illustrated below.
Therefore, (753)9 – (476)9 = (266)9
Converting number from base x to base y, none of x and y is equal to 10
The standard approach is to convert the number in base x to the number in base 10 and then again convert this number in base 10 to the number in base y. i.e.
(number)x ————–> (number)10 —————> (number)y
The process is repetitive and tedious if the number is large. However if the base y is some power of x, then the conversion is easy.
Example: Convert (10011110101)2 to base 4.
Solution:
Note that 22 = 4, also the digits used in base 4 are 0, 1, 2, and 3.
The conversions of these digits in base 4 to base 2 are as follow:
(0)4 = (00)2
(1)4 = (01)2
(2)4 = (10)2
(3)4 = (10)2
To convert (10011110101)2 into base 4, we write the digits in pairs starting from the right hand side. i.e.
(10011110101)2 = (01 00 11 11 01 01)2
Now substituting the above conversion from base 2 to base 4 we get
(10011110101)2 = (1 0 3 3 1 1)4
Similarly, if we have to convert (2322310)4 into base 2, it can easily be done by substitution as we did in the above problem, i.e.
(2322310)4 = (10 11 10 10 11 01 00)2
The logic can be extended to base 8 also by using the following substitution:
| Binary | Octal |
| 000 | 0 |
| 001 | 1 |
| 010 | 2 |
| 011 | 3 |
| 100 | 4 |
| 101 | 5 |
| 110 | 6 |
| 111 | 7 |
Example: convert (100010101010100011110)2 into base 8.
Solution:
Grouping digits of the number in groups of three from the right hand side we get (100 010 101 010 100 011 110)2
Now substituting each group of digits with the corresponding equivalence in base 8 we get:
(100 010 101 010 100 011 110)2 = (4 2 5 2 4 3 6)8
Important results related to divisibility rules in different bases
- If the sum of all the digits of a number in base x is divisible by x-1, then the number itself is divisible by x-1.
Example: Is (343626)9 divisible by 8?
Solution:
The sum of digits = 3 + 4 + 3 + 6 + 2 + 6 = 24. As 24 is divisible by 8, the number (343626)9 is also divisible by 8.
- If a number in base x has even number of digits, and the number is a palindrome i.e. the digits equidistant from each end are the same, then the number is divisible by x+1.
Example: (234432)7 is a six digit palindrome number and by the above result, it is divisible by 7+1 = 8.
- Number system syllabus and preparation tips for CAT exam
- HCF and LCM Concepts
- Number of Factors of a Number
- Concepts of Remainders
- Remainders - Fermat Theorem
- Chinese Remainder Theorem
- 3 Steps to Find Last Two Digit Numbers
- Factorials - Number of Trailing Zeros
- Base System Concepts and Questions
