7) Given a table with 2 rows and 1007 columns, let's write all the numbers from 1 to 1007 on the first row in ascending order, and the numbers from 1008 to 2014 on the second row, also in ascending order. Now, let's look at the table as 1007 pairs of numbers stacked vertically: in how many of them is the number in the second row a multiple of the one above it?
(A) 2
(B)
(C) 4
(D)
(E) 6
7. The answer is $\mathbf{( C )}$.

The pairs are of the form $(k, k+1007)$. We know that $k$ divides $k+1007$ if and only if $k$ divides 1007. Since $1007=19 \cdot 53$, and since the numbers 19 and 53 are prime, the only possible $k$ values are $k=1,19,53,1007$.

[Problem proposed by G. Barbarino]