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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 1010 and then convert the number to the desired base.

Example 5

The number 23422342 in base 66 when written in the base 99 system is

Solution

We first convert the number to base 1010.
(2342)6=(2×63)+(3×62)+(4×6)+2(2342)_{6} = (2 \times 6^{3}) + (3 \times 6^{2}) + (4 \times 6) + 2

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

Now, we successively divide 566566 by 99.



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

Answer: 688688


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 aa and bb are the two bases and an=ba^{n} = b, then each digit of the number in base bb has nn digits in base aa.

Example 6

If the number 156156 in base 88 is xx in base 22, then x=x = ?

Solution

The two bases 22 and 88 are linked as 23=82^{3} = 8.
∴ Each digit of the number in base 88 is written as 33 digits in base 22.

Base 22 can have the digits 00 and 11 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}
5=(1×22)+(0×21)+(1×20)=(101)25 = (1 \times 2^{2}) + (0 \times 2^{1}) + (1 \times 2^{0}) = (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}

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

Answer: 11011101101110


Example 7

21222012122201 in base 33 when written in base 99 is

Solution

The two bases are 33 and 32=93^{2} = 9.

Every 22 digits from the right in the base 33 number can be combined and represented as one digit in base 99.

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

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

Answer: 25812581


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