7. How many positive integers $n$ exist such that both $\frac{n+1}{3}$ and $3 n+1$ are three-digit integers?
Solution
012

For $\frac{n+1}{3}$ to be a three-figure integer we require $99<\frac{n+1}{3}<999$.
This simplifies to $297<n+1<2997$, that is $296<n<2996$.
For $3 n+1$ to be a three-figure integer we require $99<3 n+1<999$.
This simplifies to $98<3 n<998$, that is $\frac{98}{3}<n<\frac{998}{3}$.
These inequalities are simultaneously solved when $296<n<\frac{998}{3}=332 \frac{2}{3}$.
For every integer value of $n$ between 297 and 332 it is clear that $3 n+1$ will be a three-figure integer. However, $\frac{n+1}{3}$ will only be an integer for those values of $n+1$ which are divisible by 3 . These are $299,302,305,308,311,314,317,320,323,326,329$ and 332 . There are 12 numbers in this list.