Converting a Number from One Base to Another Base
To convert a number given in any base to a different target base, the number is first expanded to find its equivalent in base 10. Then, this base 10 value is converted to the desired target base using the repeated division method.
Examples:
- Let’s find the equivalent of the binary number $(1011101)_2$ in base 3.
First, we convert it to base 10:
\[
\begin{aligned}
(1011101)_2 &= 1\cdot2^6+0\cdot2^5+1\cdot2^4+1\cdot2^3+1\cdot2^2+0\cdot2^1+1\cdot2^0 \\
&= 64+16+8+4+1 = 93
\end{aligned}
\]
So, $(1011101)_2 = 93_{10}$. Now, we convert 93 to base 3 using successive divisions:
\[
\begin{array}{l}
1. \quad 93 \div 3 = 31 \quad \text{remainder} = 0 \\
2. \quad 31 \div 3 = 10 \quad \text{remainder} = 1 \\
3. \quad 10 \div 3 = 3 \quad \text{remainder} = 1 \\
4. \quad 3 \div 3 = 1 \quad \text{remainder} = 0 \\
5. \quad 1 \div 3 = 0 \quad \text{remainder} = 1
\end{array}
\]
Reading the remainders from bottom to top, we get $(10110)_3$. Thus, $(1011101)_2 = 93_{10} = (10110)_3$.
- Let’s find the equivalent of the number $(123)_4$ in base 5.
First, we convert it to base 10:
\[
\begin{aligned}
(123)_4 &= 1\cdot 4^2 + 2\cdot 4^1 + 3\cdot 4^0 \\
&= 16 + 8 + 3 = 27
\end{aligned}
\]
So, $(123)_4 = 27_{10}$. Now, we convert 27 to base 5:
\[
\begin{array}{l}
1. \quad 27 \div 5 = 5 \quad \text{remainder} = 2 \\
2. \quad 5 \div 5 = 1 \quad \text{remainder} = 0 \\
3. \quad 1 \div 5 = 0 \quad \text{remainder} = 1
\end{array}
\]
Reading the remainders backwards, we get $27 = (102)_5$. Therefore, $(123)_4 = 27_{10} = (102)_5$.
Converting a Number from Base $a$ to Base $a^n$
When short-cutting the conversion of a number from base $a$ directly to base $a^n$:
- The digits of the given number are grouped into sets of $n$ digits, working from right to left.
- Each group is independently evaluated as a separate positional value in base $a$ to find the corresponding single digit in base $a^n$.
- These calculated digits are written down in their respective order from left to right to form the final number in base $a^n$.
Examples:
- Let’s find the equivalent of $(10110011010)_2$ in base 8.
Since $8 = 2^3$, we separate the binary digits into groups of 3 from right to left. Pad with leading zeros on the far left if necessary:
\[
\begin{array}{l}
(010 \quad 110 \quad 011 \quad 010)_2 = (abcd)_8 \\
(010)_2 = 0\cdot2^2 + 1\cdot2^1 + 0\cdot2^0 = 2 = a \\
(110)_2 = 1\cdot2^2 + 1\cdot2^1 + 0\cdot2^0 = 6 = b \\
(011)_2 = 0\cdot2^2 + 1\cdot2^1 + 1\cdot2^0 = 3 = c \\
(010)_2 = 0\cdot2^2 + 1\cdot2^1 + 0\cdot2^0 = 2 = d
\end{array}
\]
Combining these digits yields: $(10110011010)_2 = (2632)_8$
- Let’s find the equivalent of $(2012201)_3$ in base 9
Since $9 = 3^2$, we separate the digits into groups of 2 from right to left:
\[
\begin{array}{l}
(02 \quad 01 \quad 22 \quad 01)_3 = (abcd)_9 \\
(02)_3 = 2 = a \\
(01)_3 = 1 = b \\
(22)_3 = 2\cdot3^1 + 2\cdot3^0 = 8 = c \\
(01)_3 = 1 = d
\end{array}
\]
Combining these digits yields: $(2012201)_3 = (2181)_9$
- Let’s find the equivalent of $(101010100111010101)_2$ in base 16
Since $16 = 2^4$, we partition the binary string into chunks of 4 digits from right to left:
\[
\begin{array}{l}
(10 \quad 1010 \quad 1001 \quad 1101 \quad 0101)_2 = (abcde)_{16} \\
(0010)_2 = 2 = a \\
(1010)_2 = 10 \longrightarrow A_{16} = b \\
(1001)_2 = 9 = c \\
(1101)_2 = 13 \longrightarrow D_{16} = d \\
(0105)_2 = 5 = e
\end{array}
\]
Combining these alphanumeric digits yields: $(101010100111010101)_2 = (2A9D5)_{16}$
Question 27:
What is the equivalent in base 4 of the largest four-digit odd number in base 5 that has all distinct digits?
\[ \text{A) } (2203)_4 \quad \text{B) } (3011)_4 \quad \text{C) } (3201)_4 \quad \text{D) } (21021)_4 \quad \text{E) } (20121)_4 \]
Solution:
Let the target number in base 5 be $(abcd)_5$. To maximize it with distinct digits, we assign the largest possible values starting from the highest place value: $a=4, b=3, c=2$. For a number to be odd in an odd base (like base 5), the sum of its digits must be odd, or alternatively, the last digit must be odd. The remaining available odd digit that is distinct is $1$. Thus, the maximum distinct-digit odd number is $(4321)_5$.
Let’s convert $(4321)_5$ to base 10:
\[ (4321)_5 = 4\cdot 5^3 + 3\cdot 5^2 + 2\cdot 5^1 + 1\cdot 5^0 = 500 + 75 + 10 + 1 = 586 \]
Now, we convert 586 to base 4 via successive divisions:
\[
\begin{array}{l}
586 \div 4 = 146 \quad \text{remainder } 2 \\
146 \div 4 = 36 \quad \text{remainder } 2 \\
36 \div 4 = 9 \quad \text{remainder } 0 \\
9 \div 4 = 2 \quad \text{remainder } 1 \\
2 \div 4 = 0 \quad \text{remainder } 2
\end{array}
\]
Writing the remainders in reverse order yields: $586_{10} = (21022)_4$.
Correct Answer: D
Question 28:
What is the equivalent of the hexadecimal number $(5A2)_{16}$ in base 2 ?
\[ \text{A) } (101101010)_2 \quad \text{B) } (10110100010)_2 \quad \text{C) } (111010010)_2 \quad \text{D) } (1010101011)_2 \quad \text{E) } (1100110101)_2 \]
Solution:
Since $16 = 2^4$, each hexadecimal digit maps directly to a 4-digit binary block. We expand each digit from left to right:
\[
\begin{array}{l}
5 = (0101)_2 \\
A = 10 = (1010)_2 \\
2 = (0010)_2
\end{array}
\]
Stringing these groups together gives $(010110100010)_2$. Eliminating the insignificant leading zero on the far left results in $(10110100010)_2$.
Correct Answer: B
Question 29:
How many digits are there in the binary (base 2) equivalent of the hexadecimal number $(5A93C2)_{16}$ ?
\[ \text{A) } 16 \quad \text{B) } 21 \quad \text{C) } 23 \quad \text{D) } 24 \quad \text{E) } 26 \]
Solution:
Because $16 = 2^4$, each hexadecimal digit yields exactly 4 binary bits, except potentially the leading non-zero hexadecimal digit on the far left.
The number $(5A93C2)_{16}$ has 6 digits. The 5 lower-order digits ($A, 9, 3, C, 2$) provide exactly $5 \times 4 = 20$ bits. The leading digit $5$ is represented in binary as $(101)_2$, which requires 3 bits.
Therefore, the total number of digits is $20 + 3 = 23$.
Correct Answer: C