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Math Help - Number Theory 2

  1. #1
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    Number Theory 2

    when i distribute a packet of chocolate to 7 students,i am left with 4 chocolates.when i distribute the same packet of chocolate to 11 students
    i am left with 6 chocolates . How many chocolates will be left with me if i distribute the same packet of chocolate among 13 students
    ( a packet of chocolate contains total number of chocolates 1000 < N <2000) ??
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  2. #2
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    Re: Number Theory 2

    You can use the Chinese remainder theorem.

    However your range of values is too large to allow a unique solution.

    For example

    1040 \equiv 1117\equiv 4 \mod 7

    1040\equiv 1117 \equiv 6 \mod 11

    but

    1040\equiv 0 \mod 13 and 1117\equiv 12 \mod 13.
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  3. #3
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    Re: Number Theory 2

    Hello, timkuc!

    When I distribute a packet of chocolate to 7 students, I am left with 4 chocolates.
    When I distribute the same packet of chocolate to 11 students, I am left with 6 chocolates.
    How many chocolates will be left with me if i distribute the same packet of chocolate among 13 students?
    (A packet of chocolate contains total number of chocolates 1000 < N < 2000)

    This last statement gives us over a dozen possible solutions.

    Let N = number of chocolates in a packet.

    We are told: . \begin{Bmatrix}N &=& 7a + 4 \\ N &=& 11b + 6\end{Bmatrix}\;\text{ for some positive integers }a\text{ and }b.

    Then: . 7a + 4 \:=\:11b + 6 \quad\Rightarrow\quad a \:=\:\frac{11b+2}{7} \quad\Rightarrow\quad a \:=\:b + \frac{4b+2}{7} .[1]

    Since a is an integer, 4b+2 must be a multiple of 7.

    Hence, 4b+2 \:=\:7c\;\text{ for some integer }c.

    Then: . b \:=\:\frac{7c-2}{4} \quad\Rightarrow\quad b \:=\:c + \frac{3c-2}{4} .[2]

    Since b is an integer, 3c-2 must be a multiple of 4.

    This happens when c \:=\:4k-2\;\text{ for some integer }k.


    Substitute into [2]: . b \:=\:(4k+2) + \frac{3(4k+2)-2}{4} \quad\Rightarrow\quad b \:=\:7k-3

    Substitute into [1]: . a \:=\:(7k+3) + \frac{4(7k+3) + 2}{7} \quad\Rightarrow\quad a \:=\:11k+5

    Hence: . N \:=\:7a+4 \:=\:7(11k+5) + 4 \quad\Rightarrow\quad N \:=\:77k + 39


    Since N is between 1000 and 2000:

    . . 1000\;<\;77k + 39 \;<\;2000

    . . . . . 961\;<\;77k \;<\;1961

    . . . . . . \frac{961}{77} \;<\;k \;<\;\frac{1961}{77}

    . . . . . . . 13 \:\le\:k\:\le\:25


    Therefore, there are 13 possible values for N,
    . . ranging from 1040 to 1964 in increments of 77.

    Thanks from a tutor and timkuc
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  4. #4
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    Re: Number Theory 2

    I assume you mean that when you "distribute" the chocolates you give an equal number to each students.
    "when i distribute a packet of chocolate to 7 students,i am left with 4 chocolates."
    Let X be the total number of chocolates. If you give "n" chocolates to each student, then 7n+ 4= X.

    "when i distribute the same packet of chocolate to 11 students i am left with 6 chocolates ."
    If you give "m" chocolates to each student then 11m+ 6= X.

    That is, X= 7n+ 4= 11m+ 6 so that 7m- 11n= 2. Since m and n must be integers, that is a "Diophantine equation".

    7 divides into 11 once with remainder 4: 11(1)- 7(1)= 4. 4 divides into 7 once with remainder 3: 7(1)- 4(1)= 3. 3 divides into 4 once with remaider 1: 4(1)- 3(1)= 1. Replacing the "3" with 7(1)- 4(1), 4(1)- (7(1)- 4(1))= 4(2)- 7(1)= 1. Replacing that "4" with 11(1)- 7(1), (11(1)- 7(1))(2)- 7(1)= 11(2)- 7(3)= 1. Multiplying both sides by 2, 11(4)- 7(6)= 2.

    That tells us that one solution to 7m- 11n= 2 is n= -4 and m= -6. Of course, those are not solutions to this problem because they are negative numbers. But it is easy to see that n= -4+ 7k, m= -6+ 11k is also a solution, for any k: 7(-6+ 11k)- 11(-4+ 7k)= -42+ 44+ 77k- 77k= 2 for all k.

    With n= -4+ 7k, the first equation, 7n+ 4= X, becomes 7(-4+ 7k)+ 4= 49k- 24= X. Now find k so that X, the total number of chocolates, is between 1000 and 2000.
    Thanks from a tutor and timkuc
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  5. #5
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    Re: Number Theory 2

    They look like fun methods.


    Using the Chinese remainder theorem:

    n\equiv 4 \mod 7

    and

    n\equiv 6 \mod 11

    Now,

    2\times 11 \equiv 1 \mod 7

    and

    8\times 7 \equiv 1 \mod 11

    So

    n\equiv 4\times 2 \times 11 + 6 \times 8 \times 7 \mod 7

    and

    n\equiv 4\times 2 \times 11 + 6 \times 8 \times 7 \mod 11

    So

    n\equiv 424 \equiv 39 \mod 77
    Thanks from timkuc
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  6. #6
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    Re: Number Theory 2

    soroban ,
    HallsofIvy , a tutor


    You guys are AWESOME ...

    Thanks for help
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