The number of positive integers $p$ for which $-1<\sqrt{p}-\sqrt{100}<1$ is
(A) 19
(B) 21
(C) 38
(D) 39
(E) 41
If $-1<\sqrt{p}-\sqrt{100}<1$, then $-1<\sqrt{p}-10<1$ or $9<\sqrt{p}<11$.

Since $\sqrt{p}$ is greater than 9 , then $p$ is greater than $9^{2}=81$.

Since $\sqrt{p}$ is less than 11 , then $p$ is less than $11^{2}=121$.

In other words, $81<p<121$.

Since $p$ is a positive integer, then $82 \leq p \leq 120$.

Therefore, there are $120-82+1=39$ such integers $p$.

ANSWER: (D)