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Convert the following numbers between number systems: 1) 630 to base 10, 2) DXIX to Hindu-Arabic numerals, 3) 496 to Roman numerals, 4) 111 base two to base ten.

Mathematics

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Step-by-step solution

24 steps

  1. Calculate the product of 6×1006 \times 100.

    6×100=6006 \times 100 = 600

  2. Calculate the product of 3×103 \times 10.

    3×10=303 \times 10 = 30

  3. Add the results from Step 1 and Step 2.

    600+30=630600 + 30 = 630

    The number is 630\boxed{\text{630}}.

    10. Write DXIX in Hindu Arabic numerals.

  4. Identify the value of each Roman numeral.

    D represents 500.
    X represents 10.
    I represents 1.

  5. Combine the numerals, noting that IX is a subtractive notation.

    IX means 10−1=910 - 1 = 9.

  6. Add the values.

    DXIX=D+IX=500+9=509\text{DXIX} = \text{D} + \text{IX} = 500 + 9 = 509

    The Hindu Arabic numeral is 509\boxed{\text{509}}.

    11. Write 496 in Roman numerals.

  7. Break down the number 496 into its place values for Roman numeral conversion.

    400 is represented as CD (500 - 100).
    90 is represented as XC (100 - 10).
    6 is represented as VI (5 + 1).

  8. Combine the Roman numeral representations.

    496=CD+XC+VI=CDXCVI\text{496} = \text{CD} + \text{XC} + \text{VI} = \text{CDXCVI}

    The Roman numeral is CDXCVI\boxed{\text{CDXCVI}}.

    12. Convert 111two111_{\text{two}} to base ten.

  9. Write the binary number in expanded form using powers of 2.

    111two=(1×22)+(1×21)+(1×20)111_{\text{two}} = (1 \times 2^2) + (1 \times 2^1) + (1 \times 2^0)

  10. Calculate the powers of 2.

    22=42^2 = 4 21=22^1 = 2 20=12^0 = 1

  11. Substitute the power values and sum the terms.

    111two=(1×4)+(1×2)+(1×1)=4+2+1=7111_{\text{two}} = (1 \times 4) + (1 \times 2) + (1 \times 1) = 4 + 2 + 1 = 7

    The base ten equivalent is 7\boxed{\text{7}}.

    13. Convert 23ten23_{\text{ten}} to binary.

  12. Divide 23 by 2 and record the remainder.

    23÷2=11 remainder 123 \div 2 = 11 \text{ remainder } 1

  13. Divide the quotient (11) by 2 and record the remainder.

    11÷2=5 remainder 111 \div 2 = 5 \text{ remainder } 1

  14. Continue dividing the quotient by 2 until the quotient is 0.

    5÷2=2 remainder 15 \div 2 = 2 \text{ remainder } 1 2÷2=1 remainder 02 \div 2 = 1 \text{ remainder } 0 1÷2=0 remainder 11 \div 2 = 0 \text{ remainder } 1

  15. Read the remainders from bottom to top to get the binary number.

    23ten=10111two23_{\text{ten}} = \boxed{10111_{\text{two}}}

    14. If 23n=15ten23_n = 15_{\text{ten}}. Find n.

  16. Convert the number 23n23_n from base nn to base 10.

    23n=(2×n1)+(3×n0)=2n+323_n = (2 \times n^1) + (3 \times n^0) = 2n + 3

  17. Set up the equation by equating the base 10 representation to 15ten15_{\text{ten}}.

    2n+3=152n + 3 = 15

  18. Solve the linear equation for nn.

    2n=15−32n = 15 - 3 2n=122n = 12 n=122n = \frac{12}{2} n=6n = 6

    The value of n is 6\boxed{\text{6}}.

    15. Add: 101two+11two101_{\text{two}} + 11_{\text{two}}

  19. Add the rightmost digits (least significant bits).

    1two+1two=10two1_{\text{two}} + 1_{\text{two}} = 10_{\text{two}}. Write down 0, carry over 1.

  20. Add the next column of digits, including the carry-over.

    0two+1two+1carry=10two0_{\text{two}} + 1_{\text{two}} + 1_{\text{carry}} = 10_{\text{two}}. Write down 0, carry over 1.

  21. Add the leftmost digits, including the carry-over.

    1two+0two+1carry=10two1_{\text{two}} + 0_{\text{two}} + 1_{\text{carry}} = 10_{\text{two}}. Write down 10.

    1 0 1two+1 1two1 0 0 0two\begin{align*} &\quad 1 \ 0 \ 1_{\text{two}} \\ + &\quad \quad 1 \ 1_{\text{two}} \\ \hline &\quad 1 \ 0 \ 0 \ 0_{\text{two}}\end{align*}

    The sum is 1000two\boxed{1000_{\text{two}}}.

    16. Subtract: 100two−11two100_{\text{two}} - 11_{\text{two}}

  22. Set up the subtraction.

    1 0 0two−1 1two\begin{align*} &\quad 1 \ 0 \ 0_{\text{two}} \\ - &\quad \quad 1 \ 1_{\text{two}} \\ \hline \end{align*}

  23. Subtract the rightmost digits.

    0−10 - 1 requires borrowing. Borrow from the 212^1 place, which is also 0. So, borrow from the 222^2 place.
    The 1 in the 222^2 place becomes 0. The 0 in the 212^1 place becomes 10two10_{\text{two}} (which is 2ten2_{\text{ten}}).
    Now, borrow from the 10two10_{\text{two}} in the 212^1 place. It becomes 1two1_{\text{two}}. The 0 in the 202^0 place becomes 10two10_{\text{two}}.

    10 01 0two0 101 10two−1 1two\begin{align*} &\quad \cancel{1}^0 \ \cancel{0}^{1} \ 0_{\text{two}} \\ &\quad 0 \ \cancel{10}^1 \ 10_{\text{two}} \\ - &\quad \quad \quad 1 \ 1_{\text{two}} \\ \hline \end{align*}

  24. Perform the subtraction column by column.

    202^0 column: 10two−1two=1two10_{\text{two}} - 1_{\text{two}} = 1_{\text{two}}.
    212^1 column: 1two−1two=0two1_{\text{two}} - 1_{\text{two}} = 0_{\text{two}}.
    222^2 column: 0two−0two=0two0_{\text{two}} - 0_{\text{two}} = 0_{\text{two}}.

    1 0 0two−1 1two1two\begin{align*} &\quad 1 \ 0 \ 0_{\text{two}} \\ - &\quad \quad 1 \ 1_{\text{two}} \\ \hline &\quad \quad \quad 1_{\text{two}}\end{align*}

    The difference is 1two\boxed{1_{\text{two}}}.

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