Converting a Number from Any Base to Base 10

The equivalent of a number in any base can be found in base 10 by expanding it according to its place values.

Let’s find the base 10 equivalents of the numbers $(118)_9$, $(1034)_5$, $(21203)_4$, and $(101101)_2$.

\[ (118)_9 = 1\cdot 9^2+ 1\cdot 9^1 + 8\cdot 9^0 =98 \]

\[ (1034)_5 = 1\cdot 5^3+ 0\cdot 5^2 + 3\cdot 5^1 + 4\cdot 5^0 =144 \]

\[ (21203)_4 = 2\cdot 4^4+ 1\cdot 4^3 + 2\cdot 4^2+0\cdot 4^1+3\cdot 4^0 =611 \]

\[ (101101)_2 = 1\cdot 2^5 + 0\cdot 2^4 + 1\cdot 2^3 + 1\cdot 2^2+0\cdot 2^1+1\cdot 2^0= 45 \]

Let’s find the base 10 equivalents of the numbers $(110.01)_2$ and $(123.21)_4$.

\[ (110.01)_2 = 1\cdot 2^2+ 1\cdot 2^1 + 0\cdot 2^0 + 0\cdot 2^{-1}+ 1\cdot 2^{-2}= 4+2+\dfrac{1}{4}=\dfrac{25}{4}=6.25 \]

\[ (123.21)_4 = 1\cdot 4^2+ 2\cdot 4^1 + 3\cdot 4^0 + 2\cdot 4^{-1}+ 1\cdot 4^{-2}= 27+\dfrac{2}{4}+\dfrac{1}{16}= 27.5625 \]

Question 24:

Given that $x$ represents the number base and $(132)_x=42$, what is the value of $x$?

\[\text{A) 8 \quad B) 7 \quad C) 6 \quad D) 5 \quad E) 4}\]

Solution:

\[
\begin{array}{l}
(132)_x=42\\
\Rightarrow 1\cdot x^2 + 3\cdot x^1 + 2\cdot x^0 =42\\
\Rightarrow x^2 + 3x – 40 = 0\\
\Rightarrow (x-5)\cdot(x+8)=0\\
\Rightarrow x-5=0 \quad \text{or} \quad x+8 = 0 \\
\Rightarrow x= 5 \quad \text{or} \quad x=-8\\
\text{Since } x \text{ is a number base, } x=5.
\end{array}
\]

Correct Answer: D

Question 25:

Given that $x$ represents the number base and $(43)_x+(36)_x=(101)_x$, what is the value of $x$?

\[\text{A) 7 \quad B) 8 \quad C) 9 \quad D) 10 \quad E) 11}\]

Solution:

\[
\begin{array}{l}
\Rightarrow (4x^1+3) + (3x^1+6)= (x^2+x^1\cdot0+1)\\
\Rightarrow x^2-7x-8=0\\
\Rightarrow (x-8)\cdot (x+1)=0\\
\text{From here, } x=8 \quad \text{or} \quad x=-1\\
\text{Since } x \text{ is a number base, it must be } x = 8.
\end{array}
\]

Correct Answer: B

Parity (Even and Odd Numbers) in Numeral Systems

For an $(n + 1)$-digit number $R = (r_n r_{n-1} \dots r_2 r_1 r_0)_a$ in base $a$:

If the base $a$ is even: $R$ is even if the last digit $r_0$ is even; $R$ is odd if $r_0$ is odd.
If the base $a$ is odd: $R$ is odd if the sum of all its digits $(r_0 + r_1 + r_2 + \dots + r_n)$ is odd; $R$ is even if the sum is even.

Examples:

The numbers $(10010)_2$, $(3214)_6$, and $(12302)_4$ are even.

The numbers $(101001)_2$, $(43213)_6$, and $(30221)_4$ are odd.

Since $1+2+1+4 = 8$ (an even number), the number $(1214)_5$ is even. Since $1 + 0 + 2 = 3$ (an odd number), the number $(102)_3$ is odd.

Converting a Number from Base 10 to Any Base

To convert a number given in base 10 into a target base, the number is repeatedly divided by the target base. If the resulting quotient is not smaller than the target base, it is divided again. This process continues successively until a quotient smaller than the target base is obtained.

Once a quotient smaller than the target base is reached, this final quotient and the remainders obtained from the previous divisions are written in reverse order (from the last step to the first) to form the digits of the number in the target base.

Example:

Let’s find the equivalent of the number 124 in base 4. Result: $124 = (1330)_4$

\[
\begin{array}{l}
\text{Step 1: } 124 \div 4 = 31 \quad \text{Remainder: } 0 \\
\text{Step 2: } 31 \div 4 = 7 \quad \text{Remainder: } 3 \\
\text{Step 3: } 7 \div 4 = 1 \quad \text{Remainder: } 3 \\
\text{Step 4: } 1 \div 4 = 0 \quad \text{Remainder: } 1 \\
\text{Writing the remainders from the last step backwards to the first:} \\
\text{Result: } 124_{10} = (1330)_4
\end{array}
\]

Example:

Let’s find the equivalent of the number 11 in base 2.

\[
\begin{array}{l}
\text{Step 1: } 11 \div 2 = 5 \quad \text{Remainder: } 1 \\
\text{Step 2: } 5 \div 2 = 2 \quad \text{Remainder: } 1 \\
\text{Step 3: } 2 \div 2 = 1 \quad \text{Remainder: } 0 \\
\text{Step 4: } 1 \div 2 = 0 \quad \text{Remainder: } 1 \\
\text{Writing the remainders from the last step backwards to the first:} \\
\text{Result: } 11_{10} = (1011)_2
\end{array}
\]

Question 26:

The representation of the number 3975 in base 11 is $(abcd)_{11}$. Accordingly, what is the equivalent of the sum $a + b + c + d$ in base 11?

\[ \text{A) } (25)_{11} \quad \text{B) } (23)_{11} \quad \text{C) } (52)_{11} \quad \text{D) } (32)_{11} \quad \text{E) } (A1)_{11} \]

Solution:

\[
\begin{array}{l}
\text{Step 1: Convert the number 3975 to base 11:}\\
3975 \div 11 = 361 \quad \text{Remainder: } 4 \quad (d = 4) \\
361 \div 11 = 32 \quad \text{Remainder: } 9 \quad (c = 9) \\
32 \div 11 = 2 \quad \text{Remainder: } 10 \quad (b = A) \\
2 \div 11 = 0 \quad \text{Remainder: } 2 \quad (a = 2) \\
\text{As a result of these operations:} \\
3975_{10} = (2A94)_{11}\\
\\
\text{Step 2: Find the sum } a + b + c + d:\\
a + b + c + d = 2 + 10 + 9 + 4 = 25\\
\\
\text{Step 3: Convert this sum into base 11:}\\
25 \div 11 = 2 \quad \text{Remainder: } 3 \\
\text{Result: } 25_{10} = (23)_{11}
\end{array}
\]

Correct Answer: B