1. ## 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) ??

2. ## 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

$\displaystyle 1040 \equiv 1117\equiv 4 \mod 7$

$\displaystyle 1040\equiv 1117 \equiv 6 \mod 11$

but

$\displaystyle 1040\equiv 0 \mod 13$ and $\displaystyle 1117\equiv 12 \mod 13$.

3. ## 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 $\displaystyle N$ = number of chocolates in a packet.

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

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

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

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

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

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

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

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

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

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

Since $\displaystyle N$ is between 1000 and 2000:

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

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

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

. . . . . . . $\displaystyle 13 \:\le\:k\:\le\:25$

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

4. ## 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.

5. ## Re: Number Theory 2

They look like fun methods.

Using the Chinese remainder theorem:

$\displaystyle n\equiv 4 \mod 7$

and

$\displaystyle n\equiv 6 \mod 11$

Now,

$\displaystyle 2\times 11 \equiv 1 \mod 7$

and

$\displaystyle 8\times 7 \equiv 1 \mod 11$

So

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

and

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

So

$\displaystyle n\equiv 424 \equiv 39 \mod 77$

6. ## Re: Number Theory 2

soroban ,
HallsofIvy , a tutor

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# 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 left with me if I distribute the same packet of chocol

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