12 positive integers are written in a row. The fourth number is 4, and the 12 number is 12. The sum of any three neighboring numbers is 333.Determine the twelve integers
You have the numbers: $\displaystyle A,B,C,4,D,E,F,G,H,I,J,12$
You know that: $\displaystyle 4+C+D=333$
Solve to get: $\displaystyle C=329-D$
You also know that: $\displaystyle 4+D+E=333$
Solve to get: $\displaystyle E=329-D$
Therefore: $\displaystyle C=E$
Using the same method you get that: $\displaystyle B=D$
So you know that: $\displaystyle C+4+D=333$
Substitute: $\displaystyle C+4+B=333$
Solve: $\displaystyle 4=333-B-C$
Note that: $\displaystyle A+B+C=333$
Solve: $\displaystyle A=333-B-C$
Therefore: $\displaystyle A=4$
Using the same methods above you'll find that:
$\displaystyle A=4=F=I$
$\displaystyle B=D=G=J$
$\displaystyle C=E=H=12$
Remember that: $\displaystyle D=333-4-C$
Substitute: $\displaystyle D=333-4-12$
Then: $\displaystyle D=317$
So the pattern is: $\displaystyle 4,317,12,4,317,12,4,317,12,4,317,12$