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Remainder Puzzle I (math04)
Problem: Is there a number which gives a remainder of 1 when divided by 3, and a
remainder of 2 when divided by 4, and a remainder of 3 when divided by 5 and nally a
remainder of 4 when divided by 6? Find the smallest such number, if any exists. (see [2]
and [3]).
Modeling Steps
The unknown number when divided by 3, 4, 5, 6 leaves a remainder of 1, 2, 3, 4.
1. Let the unknown integer number be N .
2. We introduce an aditional 4 integer variables x1 , x2 , x3 , and x4 representing the
results of the divisions without remainder.
3. If N divided by three leaves a remainder of 1, then x1 = (N − 1)/3 must be an
integer number.
4. If N divided by four leaves a remainder of 2, then x2 = (N −2)/4 must be an integer
number.
5. If N divided by ve leaves a remainder of 3, then x4 = (N − 3)/5 must be an integer
number.
6. If N divided by six leaves a remainder 4, then x4 = (N − 4)/6 must be an integer
number.
7. The four requirements can be expressed as follows :
(N − i)/(i + 2) = xi ,
for all i ∈ {1 . . . 4}
8. We want to nd the smaller number N , that fulls the requirements. Therefore, N
is minimized in the objective function.
The complete model code in LPL for this model is as follows (see [1]):
Listing 1: The Model
model math04 "Remainder Puzzle I";
set i := [1..4];
integer variable x{i}; N;
constraint A{i}: (N-i)/(i+2) = x[i];
minimize obj: N;
Write(’The smallest such number N is %d\n’, N);
end
Solution: The solution is: N = 58.
The vector x is (19, 14, 11, 9). The second smallest
number which fulls the requirements is N = 118.
Question (Answer see )
1. Find the second smallest number that fulls the requirements.
1
2. Add the requirement that the number N must be divisible by 13.
3. Find the 10 smallest numbers which full the requirements.
Answer
(Question see )
1. Add the lower bound of 59 to variable N and solve again. Or, add the constraint:
constraint B: N >= 59;
The next solution found is N = 118.
2. Add a new integer variable, say M , then add the following constraint:
constraint C: N = 13*M;
The smallest such number is N = 598.
3. The simplest way to achive this is to add a loop to the model and solve the model
each time with a modied lower bound on N . (This model is implemented in the
next section math04a1 ).
References
[1] T. Hürlimann. Reference Manual for the LPL Modelling Language, most recent version. www.virtual-optima.com.
[2] B.A. Kordemsky.
[3] Clare Nolan.
html.
The Moscow Puzzles. Charles Scribner's Sons, 1972.
http://www.chlond.demon.co.uk/academic/puzzles.
1 http://lpl.unifr.ch/lpl/Solver.jsp?name=/math04a
2