Let's select 2 integers in the range of 1 TO 100, say X and Y , (X>Y).
To avoid fractions X should be a multiple of Y.
On those two integers we will perform 4 operations (addition,
subtraction, multiplication and division) and then sum up the 4 answers.
Example X = 81 and Y = 9
sum would be 90
difference 72
product 729
and quotient 9
altogether 900
It's up to the solver to convert the final result,
restoring the original values of X and Y.
Solve for final result 324
Why is anyone using a computer for a PENCIL AND PAPER problem?
For X and Y the sum of the four operations can be written as XY+(X+Y)+(X-Y)+(X/Y). Since we are given X as a multiple of Y then let X=KY.
Substituting this into the expression allows for a lot of simplification:
(KY)Y+(KY+Y)+(KY-Y)+(KY/Y)
= KY^2+2KY+K
= K*(Y+1)^2
This last expression has a clear perfect square factor. The expression is equal to 324=18^2, so we want to look for the perfect square factors, which are the squares of the factors of 18.
Over positive integers there are six factors of 18: 1, 2, 3, 6, 9, 18.
Then the corresponding factorizations of 324 are 324*1^2, 81*2^2, 36*3^2, 9*6^2, 4*9^2, and 1*18^2.
From these factorizations we get (K,Y) = (324,0), (81,1), (36,2), (9,5), (4,8), and (1,17). The first must be discarded since if Y=0 then we get division by zero in the original problem. The last must also be discarded since K=1 implies X=Y however the problem statement asks for X>Y.
From this we get four solutions (X,Y) = (81,1), (72,2), (45,5), and (32,8).
Solution
