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Number Systems

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Roman Numerals
Conversion to Base 10
Conversion to other Bases
Non-Decimal Bases
Arithmetic Operations
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CAT 2025 Lesson : Number Systems - Non-Decimal Bases

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5. Special Types

5.1 Conversion from one base to another

When two bases are not linked (explained in 5.2 Linked Bases), we convert the number to base 101010 and then convert the number to the desired base.

Example 5

The number 234223422342 in base 666 when written in the base 999 system is

Solution

We first convert the number to base 101010.
(2342)6=(2×63)+(3×62)+(4×6)+2(2342)_{6} = (2 \times 6^{3}) + (3 \times 6^{2}) + (4 \times 6) + 2(2342)6​=(2×63)+(3×62)+(4×6)+2

=432+108+24+2=566= 432 + 108 + 24 + 2 = 566=432+108+24+2=566

Now, we successively divide 566566566 by 999.



∴ (566)10=(688)9(566)_{10} = (688)_{9}(566)10​=(688)9​

Answer: 688688688


5.2 Linked bases

In conversions where one base is an integral power of another base, the conversion can be directly done without converting to the decimal system.

If aaa and bbb are the two bases and an=ba^{n} = ban=b, then each digit of the number in base bbb has nnn digits in base aaa.

Example 6

If the number 156156156 in base 888 is xxx in base 222, then x=x =x= ?

Solution

The two bases 222 and 888 are linked as 23=82^{3} = 823=8.
∴ Each digit of the number in base 888 is written as 333 digits in base 222.

Base 222 can have the digits 000 and 111 only. Instead of successively dividing, we can find the digits as follows.
1=(0×22)+(0×21)+(1×20)=(001)21 = (0 \times 2^{2}) + (0 \times 2^{1}) + (1 \times 2^{0}) = (001)_{2}1=(0×22)+(0×21)+(1×20)=(001)2​
5=(1×22)+(0×21)+(1×20)=(101)25 = (1 \times 2^{2}) + (0 \times 2^{1}) + (1 \times 2^{0}) = (101)_{2}5=(1×22)+(0×21)+(1×20)=(101)2​
6=(1×22)+(1×21)+(0×20)=(110)26 = (1 \times 2^{2}) + (1 \times 2^{1}) + (0 \times 2^{0}) = (110)_{2}6=(1×22)+(1×21)+(0×20)=(110)2​

When the above are written together, we get the number to be (0011011100)2(0011011100)_{2}(0011011100)2​ or (1101110)2(1101110)_{2}(1101110)2​

Answer: 110111011011101101110


Example 7

212220121222012122201 in base 333 when written in base 999 is

Solution

The two bases are 333 and 32=93^{2} = 932=9.

Every 222 digits from the right in the base 333 number can be combined and represented as one digit in base 999.

The number is broken as 2,12,22,012, 12, 22, 012,12,22,01.

(2)3=(2×1)=2(2)_{3} = (2 \times 1) = 2(2)3​=(2×1)=2
(12)3=(1×3)+(2×1)=5(12)_{3} = (1 \times 3) + (2 \times 1) = 5(12)3​=(1×3)+(2×1)=5
(22)3=(2×3)+(2×1)=8(22)_{3} = (2 \times 3) + (2 \times 1) = 8(22)3​=(2×3)+(2×1)=8
(01)3=(0×3)+(1×1)=1(01)_{3} = (0 \times 3) + (1 \times 1) = 1(01)3​=(0×3)+(1×1)=1

∴ (2 12 22 01)3=(2 5 8 1)9(2 \space 12 \space 22 \space 01)_{3} = (2 \space 5 \space 8 \space 1)_{9}(2 12 22 01)3​=(2 5 8 1)9​

Answer: 258125812581


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