1Convert 
11001
1
2
110011 
2
​
  to decimal:
1
∗
2
5
+
1
∗
2
4
+
0
∗
2
3
+
0
∗
2
2
+
1
∗
2
1
+
1
∗
2
0
=
51
1∗2 
5
 +1∗2 
4
 +0∗2 
3
 +0∗2 
2
 +1∗2 
1
 +1∗2 
0
 =51
2Convert 
11001
1
2
110011 
2
​
  to octal: 
11001
1
2
=
6
3
8
110011 
2
​
 =63 
8
​
 
3Convert 
34
5
10
345 
10
​
  to binary:
34
5
10
=
10101100
1
2
345 
10
​
 =101011001 
2
​
 
4Convert 
34
5
10
345 
10
​
  to hexadecimal: 
34
5
10
=
15
9
16
345 
10
​
 =159 
16
​
 
5Convert 
76
7
8
767 
8
​
  to decimal:
7
∗
8
2
+
6
∗
8
1
+
7
∗
8
0
=
511
7∗8 
2
 +6∗8 
1
 +7∗8 
0
 =511
6Convert 
76
7
8
767 
8
​
  to binary:
76
7
8
=
10000011
1
2
767 
8
​
 =100000111 
2
​
 
7Convert 
987
7
16
9877 
16
​
  to decimal:
9
∗
1
6
3
+
8
∗
1
6
2
+
7
∗
1
6
1
+
7
∗
1
6
0
=
38935
9∗16 
3
 +8∗16 
2
 +7∗16 
1
 +7∗16 
0
 =38935
8Convert 
987
7
16
9877 
16
​
  to octal: 
987
7
16
=
11044
7
8
9877 
16
​
 =110447 
8
​


Let's break down and solve each problem:

### 1. Convert \( 110011_2 \) to decimal:
\[
1 \times 2^5 + 1 \times 2^4 + 0 \times 2^3 + 0 \times 2^2 + 1 \times 2^1 + 1 \times 2^0 = 32 + 16 + 0 + 0 + 2 + 1 = 51
\]
So, \( 110011_2 = 51_{10} \).

### 2. Convert \( 110011_2 \) to octal:
First, group the binary number in sets of three from right to left:
\[
110\ 011_2
\]
Convert each group to octal:
\[
110_2 = 6_8 \quad \text{and} \quad 011_2 = 3_8
\]
So, \( 110011_2 = 63_8 \).

### 3. Convert \( 345_{10} \) to binary:
Let's divide 345 by 2 repeatedly and note the remainders:
\[
345 \div 2 = 172 \quad \text{remainder} \ 1
\]
\[
172 \div 2 = 86 \quad \text{remainder} \ 0
\]
\[
86 \div 2 = 43 \quad \text{remainder} \ 0
\]
\[
43 \div 2 = 21 \quad \text{remainder} \ 1
\]
\[
21 \div 2 = 10 \quad \text{remainder} \ 1
\]
\[
10 \div 2 = 5 \quad \text{remainder} \ 0
\]
\[
5 \div 2 = 2 \quad \text{remainder} \ 1
\]
\[
2 \div 2 = 1 \quad \text{remainder} \ 0
\]
\[
1 \div 2 = 0 \quad \text{remainder} \ 1
\]
So, reading the remainders from bottom to top, \( 345_{10} = 101011001_2 \).

### 4. Convert \( 345_{10} \) to hexadecimal:
Let's divide 345 by 16:
\[
345 \div 16 = 21 \quad \text{remainder} \ 9
\]
\[
21 \div 16 = 1 \quad \text{remainder} \ 5
\]
\[
1 \div 16 = 0 \quad \text{remainder} \ 1
\]
So, reading from bottom to top, \( 345_{10} = 159_{16} \).

### 5. Convert \( 767_8 \) to decimal:
\[
7 \times 8^2 + 6 \times 8^1 + 7 \times 8^0 = 448 + 48 + 7 = 503
\]
So, \( 767_8 = 503_{10} \).

### 6. Convert \( 767_8 \) to binary:
First, convert each octal digit to its binary equivalent:
\[
7_8 = 111_2 \quad 6_8 = 110_2 \quad 7_8 = 111_2
\]
So, \( 767_8 = 111110111_2 \).

### 7. Convert \( 9877_{16} \) to decimal:
\[
9 \times 16^3 + 8 \times 16^2 + 7 \times 16^1 + 7 \times 16^0 = 36864 + 2048 + 112 + 7 = 39031
\]
So, \( 9877_{16} = 39031_{10} \).

### 8. Convert \( 9877_{16} \) to octal:
First, convert \( 9877_{16} \) to binary:
\[
9_{16} = 1001_2 \quad 8_{16} = 1000_2 \quad 7_{16} = 0111_2 \quad 7_{16} = 0111_2
\]
So, \( 9877_{16} = 1001100001110111_2 \).

Next, group the binary number into sets of three from right to left and convert to octal:
\[
100\ 110\ 000\ 111\ 011_2 = 4\ 6\ 0\ 7\ 3_8
\]
So, \( 9877_{16} = 46073_8 \).

---

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

Here are 5 related questions for further practice:

1. Convert \( 110011_2 \) to hexadecimal.
2. Convert \( 101011001_2 \) to octal.
3. Convert \( 159_{16} \) to binary.
4. Convert \( 503_{10} \) to binary.
5. Convert \( 9877_{16} \) to binary.

**Tip:** Always double-check each step in number base conversions, especially when dealing with different number systems, to avoid mistakes.

Convert 110011 (binary) to decimal, octal, and hexadecimal; Convert 345 (decimal) to binary and hexadecimal; Convert 767 (octal) to decimal and binary; Convert 9877 (hexadecimal) to decimal and octal.

Learn how to convert numbers between different bases: binary, octal, decimal, and hexadecimal. Detailed step-by-step solutions include converting binary to decimal and octal, decimal to binary and hexadecimal, octal to decimal and binary, and hexadecimal to decimal and octal. Suitable for high school students in Grades 9-12.

Math Problem Statement

1Convert 11001 1 2 110011 2 to decimal: 1 ∗ 2 5 + 1 ∗ 2 4 + 0 ∗ 2 3 + 0 ∗ 2 2 + 1 ∗ 2 1 + 1 ∗ 2 0

51 1∗2 5 +1∗2 4 +0∗2 3 +0∗2 2 +1∗2 1 +1∗2 0 =51 2Convert 11001 1 2 110011 2 to octal: 11001 1 2

3Convert 34 5 10 345 10 to binary: 34 5 10

4Convert 34 5 10 345 10 to hexadecimal: 34 5 10

5Convert 76 7 8 767 8 to decimal: 7 ∗ 8 2 + 6 ∗ 8 1 + 7 ∗ 8 0

511 7∗8 2 +6∗8 1 +7∗8 0 =511 6Convert 76 7 8 767 8 to binary: 76 7 8

7Convert 987 7 16 9877 16 to decimal: 9 ∗ 1 6 3 + 8 ∗ 1 6 2 + 7 ∗ 1 6 1 + 7 ∗ 1 6 0

38935 9∗16 3 +8∗16 2 +7∗16 1 +7∗16 0 =38935 8Convert 987 7 16 9877 16 to octal: 987 7 16

Solution

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Math Problem Statement

1Convert 11001 1 2 110011 2 ​ to decimal: 1 ∗ 2 5 + 1 ∗ 2 4 + 0 ∗ 2 3 + 0 ∗ 2 2 + 1 ∗ 2 1 + 1 ∗ 2 0

51 1∗2 5 +1∗2 4 +0∗2 3 +0∗2 2 +1∗2 1 +1∗2 0 =51 2Convert 11001 1 2 110011 2 ​ to octal: 11001 1 2

3Convert 34 5 10 345 10 ​ to binary: 34 5 10

4Convert 34 5 10 345 10 ​ to hexadecimal: 34 5 10

5Convert 76 7 8 767 8 ​ to decimal: 7 ∗ 8 2 + 6 ∗ 8 1 + 7 ∗ 8 0

511 7∗8 2 +6∗8 1 +7∗8 0 =511 6Convert 76 7 8 767 8 ​ to binary: 76 7 8

7Convert 987 7 16 9877 16 ​ to decimal: 9 ∗ 1 6 3 + 8 ∗ 1 6 2 + 7 ∗ 1 6 1 + 7 ∗ 1 6 0

38935 9∗16 3 +8∗16 2 +7∗16 1 +7∗16 0 =38935 8Convert 987 7 16 9877 16 ​ to octal: 987 7 16