The task asks to convert the following decimal numbers into binary:

a. $1234_{10}$

b. $5670_{10}$

c. $2321_{10}$

Let's solve them one by one by converting the decimal numbers into binary:

1. 1234 in binary:

   • 1234 ÷ 2 = 617 remainder 0
   • 617 ÷ 2 = 308 remainder 1
   • 308 ÷ 2 = 154 remainder 0
   • 154 ÷ 2 = 77 remainder 0
   • 77 ÷ 2 = 38 remainder 1
   • 38 ÷ 2 = 19 remainder 0
   • 19 ÷ 2 = 9 remainder 1
   • 9 ÷ 2 = 4 remainder 1
   • 4 ÷ 2 = 2 remainder 0
   • 2 ÷ 2 = 1 remainder 0
   • 1 ÷ 2 = 0 remainder 1

   So, $1234_{10} = 10011010010_2$.

2. 5670 in binary:

   • 5670 ÷ 2 = 2835 remainder 0
   • 2835 ÷ 2 = 1417 remainder 1
   • 1417 ÷ 2 = 708 remainder 1
   • 708 ÷ 2 = 354 remainder 0
   • 354 ÷ 2 = 177 remainder 0
   • 177 ÷ 2 = 88 remainder 1
   • 88 ÷ 2 = 44 remainder 0
   • 44 ÷ 2 = 22 remainder 0
   • 22 ÷ 2 = 11 remainder 0
   • 11 ÷ 2 = 5 remainder 1
   • 5 ÷ 2 = 2 remainder 1
   • 2 ÷ 2 = 1 remainder 0
   • 1 ÷ 2 = 0 remainder 1

   So, $5670_{10} = 1011000100110_2$.

3. 2321 in binary:

   • 2321 ÷ 2 = 1160 remainder 1
   • 1160 ÷ 2 = 580 remainder 0
   • 580 ÷ 2 = 290 remainder 0
   • 290 ÷ 2 = 145 remainder 0
   • 145 ÷ 2 = 72 remainder 1
   • 72 ÷ 2 = 36 remainder 0
   • 36 ÷ 2 = 18 remainder 0
   • 18 ÷ 2 = 9 remainder 0
   • 9 ÷ 2 = 4 remainder 1
   • 4 ÷ 2 = 2 remainder 0
   • 2 ÷ 2 = 1 remainder 0
   • 1 ÷ 2 = 0 remainder 1

   So, $2321_{10} = 100100010001_2$.

Final Answers:

• $1234_{10} = 10011010010_2$
• $5670_{10} = 1011000100110_2$
• $2321_{10} = 100100010001_2$

Would you like further details or explanations on any of these conversions?

Here are 5 related questions:

1. How do you convert binary numbers back into decimal form?
2. What is the binary representation of the decimal number 999?
3. How are hexadecimal numbers related to binary?
4. How can large decimal numbers be efficiently converted to binary?
5. What are binary numbers used for in computer science?

Tip: When converting decimal to binary, divide the number by 2 repeatedly and track the remainders. Then, reverse the sequence of remainders to get the binary number.