You can use the Chinese remainder theorem.
However your range of values is too large to allow a unique solution.
For example
but
and .
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) ??
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 = number of chocolates in a packet.
We are told: .
Then: . .[1]
Since is an integer, must be a multiple of 7.
Hence,
Then: . .[2]
Since is an integer, must be a multiple of 4.
This happens when
Substitute into [2]: .
Substitute into [1]: .
Hence: .
Since is between 1000 and 2000:
. .
. . . . .
. . . . . .
. . . . . . .
Therefore, there are 13 possible values for ,
. . ranging from 1040 to 1964 in increments of 77.
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.