Since the second column includes the number 1, then step (ii) was never used on the second column, otherwise each entry would be at least 2 . To generate the 1,3 and 2 in the second column, we thus need to have used step (i) 1 time on row 1,3 times on row 2 , and 2 times on row 3 . This gives: $$\begin{array}{|l|l|l|} \hline 1 & 1 & 1 \\ \hline 3 & 3 & 3 \\ \hline 2 & 2 & 2 \\ \hline \end{array}$$ We cannot use step (i) any more times, otherwise the entries in column 2 will increase. Thus, $a=1+3+2=6$. To obtain the final grid from this current grid using only step (ii), we must increase each entry in column 1 by 6 (which means using step (ii) 3 times) and increase each entry in column 3 by 4 (which means using step (ii) 2 times). Thus, $b=3+2=5$. Therefore, $a+b=11$.