There are 7 consecutive integers and 6 consecutive integers. The average of the 7 integers is 3 less than the average of the 6 integers. The sum of the 6 integers is 6 more than the sum of the 7 integers. Find the average of the 7 integers.

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1. Let the average of the 7 integers be $m$. Then the sum of the 7 is $7m$.
2. Let the first of the 6 integers be $k$. The average of the 6 is $k+2.5$ and the sum is $6(k+2.5)$.
3. Given $m=(k+2.5)-3$ and $6(k+2.5)=7m+6$. From the first, $m=k-0.5$. Substitute into the sum equation: $6k+15=7(k-0.5)+6$.
4. Solve: $6k+15=7k+2.5$, so $k=12.5$ and $m=k-0.5=12$. Thus the average of the 7 integers is $12$.

FINAL_ANSWER: c

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