Problem:
Positive integers x and y satisfy the equation x+y=1183. What is the minimum possible value of x+y?
Answer Choices:
A. 585
B. 595
C. 623
D. 700
E. 791
Solution:
Observe that 1183=132⋅7, so 1183=137. Because x+y=137, it follows that x and y must be of the form a7 and b7, respectively, where a and b are positive integers and a+b=13. Then x=7a2 and y=7b2, so x=7a2 and y=7b2. Substituting gives
x+y=7(a2+b2)=7(a2+(13−a)2)=14a2−182a+1183
The minimum value of the quadratic polynomial occurs at a=2⋅14182=6.5. Because a and b must be positive integers, without loss of generality (by symmetry), choose a=6 and b=7. The minimum possible value of x+y is