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

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 ​

Solution

Let's break down and solve each problem:

1. Convert 1100112110011_2 to decimal:

1×25+1×24+0×23+0×22+1×21+1×20=32+16+0+0+2+1=511 \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, 1100112=5110110011_2 = 51_{10}.

2. Convert 1100112110011_2 to octal:

First, group the binary number in sets of three from right to left: 110 0112110\ 011_2 Convert each group to octal: 1102=68and0112=38110_2 = 6_8 \quad \text{and} \quad 011_2 = 3_8 So, 1100112=638110011_2 = 63_8.

3. Convert 34510345_{10} to binary:

Let's divide 345 by 2 repeatedly and note the remainders: 345÷2=172remainder 1345 \div 2 = 172 \quad \text{remainder} \ 1 172÷2=86remainder 0172 \div 2 = 86 \quad \text{remainder} \ 0 86÷2=43remainder 086 \div 2 = 43 \quad \text{remainder} \ 0 43÷2=21remainder 143 \div 2 = 21 \quad \text{remainder} \ 1 21÷2=10remainder 121 \div 2 = 10 \quad \text{remainder} \ 1 10÷2=5remainder 010 \div 2 = 5 \quad \text{remainder} \ 0 5÷2=2remainder 15 \div 2 = 2 \quad \text{remainder} \ 1 2÷2=1remainder 02 \div 2 = 1 \quad \text{remainder} \ 0 1÷2=0remainder 11 \div 2 = 0 \quad \text{remainder} \ 1 So, reading the remainders from bottom to top, 34510=1010110012345_{10} = 101011001_2.

4. Convert 34510345_{10} to hexadecimal:

Let's divide 345 by 16: 345÷16=21remainder 9345 \div 16 = 21 \quad \text{remainder} \ 9 21÷16=1remainder 521 \div 16 = 1 \quad \text{remainder} \ 5 1÷16=0remainder 11 \div 16 = 0 \quad \text{remainder} \ 1 So, reading from bottom to top, 34510=15916345_{10} = 159_{16}.

5. Convert 7678767_8 to decimal:

7×82+6×81+7×80=448+48+7=5037 \times 8^2 + 6 \times 8^1 + 7 \times 8^0 = 448 + 48 + 7 = 503 So, 7678=50310767_8 = 503_{10}.

6. Convert 7678767_8 to binary:

First, convert each octal digit to its binary equivalent: 78=111268=110278=11127_8 = 111_2 \quad 6_8 = 110_2 \quad 7_8 = 111_2 So, 7678=1111101112767_8 = 111110111_2.

7. Convert 9877169877_{16} to decimal:

9×163+8×162+7×161+7×160=36864+2048+112+7=390319 \times 16^3 + 8 \times 16^2 + 7 \times 16^1 + 7 \times 16^0 = 36864 + 2048 + 112 + 7 = 39031 So, 987716=39031109877_{16} = 39031_{10}.

8. Convert 9877169877_{16} to octal:

First, convert 9877169877_{16} to binary: 916=10012816=10002716=01112716=011129_{16} = 1001_2 \quad 8_{16} = 1000_2 \quad 7_{16} = 0111_2 \quad 7_{16} = 0111_2 So, 987716=100110000111011129877_{16} = 1001100001110111_2.

Next, group the binary number into sets of three from right to left and convert to octal: 100 110 000 111 0112=4 6 0 7 38100\ 110\ 000\ 111\ 011_2 = 4\ 6\ 0\ 7\ 3_8 So, 987716=4607389877_{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 1100112110011_2 to hexadecimal.
  2. Convert 1010110012101011001_2 to octal.
  3. Convert 15916159_{16} to binary.
  4. Convert 50310503_{10} to binary.
  5. Convert 9877169877_{16} to binary.

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

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Math Problem Analysis

Mathematical Concepts

Number Base Conversion
Binary System
Octal System
Decimal System
Hexadecimal System

Formulas

Binary to Decimal: Sum of (binary digit × 2^position)
Decimal to Binary: Repeated division by 2, reading remainders
Decimal to Hexadecimal: Repeated division by 16, reading remainders
Octal to Decimal: Sum of (octal digit × 8^position)
Hexadecimal to Decimal: Sum of (hex digit × 16^position)

Theorems

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Suitable Grade Level

Grades 9-12