In how many ways can 101 be expressed as the sum of two integers, both greater than zero, with the second integer greater than the first?
(A) 50
(B) 51
(C) 101
(D) 102
(E) 25
Beginning with the positive integer 1 as a number in the first pair, we get the sum $101=1+100$. From this point we can continue to increase the first number by one while decreasing the second number by one, keeping the sum equal to 101.

The list of possible sums is:

$$
\begin{gathered}
101=1+100 \\
101=2+99 \\
101=3+98 \\
\\
\vdots \\
101=50+51
\end{gathered}
$$

After this point, the first number will no longer be smaller than the second if we continue to add 1 to the first number and subtract 1 from the second.

There are 50 possible sums in all.

ANSWER: (A)