How many of the five numbers $101,148,200,512,621$ cannot be expressed as the sum of two or more consecutive positive integers?
(A) 0
(B) 1
(C) 2
(D) 3
(E) 4
We begin by showing that each of $101,148,200$, and 621 can be expressed as the sum of two or more consecutive positive integers.

$$
\begin{aligned}
& 101=50+51 \\
& 148=15+16+17+18+19+20+21+22 \\
& 200=38+39+40+41+42 \\
& 621=310+311
\end{aligned}
$$

We show that 512 cannot be expressed a sum of two or more consecutive positive integers. This will tell us that one of the five numbers in the list cannot be written in the desired way, and so the answer is $(\mathrm{B})$.

Now, 512 cannot be written as the sum of an odd number of consecutive positive integers.

Why is this? Suppose that 512 equals the sum of $p$ consecutive positive integers, where $p>1$ is odd.

Since $p$ is odd, then there is a middle integer $m$ in this list of $p$ integers.

Since the numbers in the list are equally spaced, then $m$ is the average of the numbers in the list.

(For example, the average of the 5 integers $6,7,8,9,10$ is 8 .)

But the sum of the integers equals the average of the integers $(m)$ times the number of integers (p). That is, $512=m p$.

Now $512=2^{9}$ and so does not have any odd divisors larger than 1 .

Therefore, 512 cannot be written as $m p$ since $m$ and $p$ are positive integers and $p>1$ is odd. Thus, 512 is not the sum of an odd number of consecutive positive integers.

Further, 512 cannot be written as the sum of an even number of consecutive positive integers. Why is this? Suppose that 512 equals the sum of $p$ consecutive positive integers, where $p>1$ is even.

Since $p$ is even, then there is not a single middle integer $m$ in this list of $p$ integers, but rather two middle integers $m$ and $m+1$.

Since the numbers in the list are equally spaced, then the average of the numbers in the list is the average of $m$ and $m+1$, or $m+\frac{1}{2}$.

(For example, the average of the 6 integers $6,7,8,9,10,11$ is $8 \frac{1}{2}$.)

But the sum of the integers equals the average of the integers $\left(m+\frac{1}{2}\right)$ times the number of integers $(p)$. That is, $512=\left(m+\frac{1}{2}\right) p$ and so $2(512)=2\left(m+\frac{1}{2}\right) p$ or $1024=(2 m+1) p$.

Now $1024=2^{10}$ and so does not have any odd divisors larger than 1 .

Therefore, 1024 cannot be written as $(2 m+1) p$ since $m$ and $p$ are positive integers and $2 m+1>1$ is odd.

Thus, 512 is not the sum of an even number of consecutive positive integers.

Therefore, 512 is not the sum of any number of consecutive positive integers.

A similar argument shows that every power of 2 cannot be written as the sum of any number of consecutive positive integers.

Returning to the original question, exactly one of the five numbers in the original list cannot be written in the desired way, and so the answer is (B).

ANsWER: (B)