Convert the hexadecimal number $20_{16}$ to decimal (base 10).
Answer:Solution. $$ 20_{16}=2\cdot 16+0\cdot 1=\boxed{32}. $$
Trivia: The whitespace character is the $20_{16}$-th character in ASCII, so URLs use %20 to encode this character.
Convert the hexadecimal number $ABC_{16}$ to decimal (base 10).
Answer:Solution. $$ \begin{align*} ABC_{16}&=A\cdot 16^2+B\cdot 16+C\\ &=10\cdot 16^2+11\cdot 16+12\\ &=2560+176+12\\ &=\boxed{2748}. \end{align*} $$
Convert $2102_{3}$ to binary.
Answer:Solution. We first convert $2102_{3}$ from base 3 to base 10:
$$ 2102_{3} = 2\cdot 3^3+1\cdot 3^2+0\cdot 3^1+2\cdot 3^0=54+9+2=65 $$Next we can either do repeated long division by $2$ to obtain the binary form of $65$, or we notice that the highest power of $2$ that goes into $65$ is $64$ so
$$ 65=64+1=2^{6}+1=\boxed{1000001_2} $$Convert $110000_{2}$ to ternary.
Answer:Solution. We first convert $110000_{2}$ from base 2 to base 10:
$$ 110000_{2} = 2^5+2^4=32+16=48 $$Next we do repeated long division by $3$ to obtain the ternary form of $48$:
$$ \begin{align*} 48&=16\cdot3+0\\ 16&=5\cdot3+1\\ 5&=1\cdot3+2\\ 1&=0\cdot3+1\\ \end{align*} $$Reading the remainders from the bottom up shows that
$$ 110000_{2}=48=\boxed{1210_3} $$Convert $121_{3}$ to base 10.
Answer:Solution.
$$ 121_{3} = 1\cdot 3^2+2\cdot 3^1+1\cdot 3^0=9+6+1=16 $$The HEX color code for the RGB color $(37, 178, 86)$ is $\#$.
Solution. We must convert each of $R$, $G$, $B$ into hexadecimal.
$$ \begin{align*} R&=37=2\cdot16+5=25_{16}\\ G&=178=11\cdot16+2=b2_{16}\\ B&=86=5\cdot16+6=56_{16}\\ \end{align*} $$Therefore the HEX code for this color is obtained by concatenating the two-digit hexadecimals we found: $\#25b256$.
Find the value of $n$ such that $79_{13}=144_{n}$.
Answer:Solution. We wish to have $7\cdot 13+9=1\cdot n^2+4n+4$ or $100=n^2+4n+4=(n+2)^2$. Taking square roots gives $\pm10=n+2$ so $n=8$ is the only positive solution.
$(11000_{2})^2$ is an $n$-digit number when written in binary. What is $n$?
Answer:Solution. We write this number out as: $$ \begin{align*} (11000_{2})^2&=(2^4+2^3)^2\\ &=(2^4)^2+2(2^4)(2^3)+(2^3)^2\\ &=2^8+2^8+2^6\\ &=2\cdot2^8+2^6\\ &=2^9+2^6\\ \end{align*} $$ So the number $(11000_{2})^2$ in binary has ten digits. ($2^9=2000000000$ in binary is a string with one 1 followed by nine 0's, for a total of ten digits).
Solution 2. Notice that $(11_{2})^2=3^2=9=1001_2$, so
$$ (11000_{2})^2=(11_{2}\times 1000_{2})^2=(11_2)^2\times 1000000_2=1001000000_2. $$This number has ten digits.
Find a digit $d$ such that $d13_{6}=1d1_{8}$.
Answer:Solution. $$ \begin{align*} &&d\cdot6^2+1\cdot 6+3&=1\cdot 8^2+d\cdot8+1\\ \Rightarrow && 36d+6+3&=64+8d+1\\ \Rightarrow && 28d&=56\\ \Rightarrow && d&=2\\ \end{align*} $$
Find a base $b$ such that $34_{b}\times21_{b}=1044_{b}$.
Answer:Solution. $$ \begin{align*} &&(3b+4)(2b+1)&=b^3+4b+4\\ \Rightarrow && 6b^2+11b+4&=b^3+4b+4\\ \Rightarrow && 0&=b^3-6b^2-7b\\ \Rightarrow && 0&=b^2-6b-7\\ \Rightarrow && 0&=(b+1)(b-7)\\ \Rightarrow && b=-1&\text{\; or\; }b=7\\ \end{align*} $$
Since we only consider positive number bases in this class, we must have $b=7$.