In the list $p, q, r, s, t, u, v, w$, each letter represents a positive integer. The sum of the values of each group of four consecutive letters in the list is 35 . If $q+v=14$, the largest possible value of $p$ is
(A) 15
(B) 19
(C) 20
(D) 23
(E) 26
Solution 1

Since the sum of $p, q, r, s$ and the sum of $q, r, s, t$ are both equal to 35 , and $q, r, s$ are added in each sum, then $p=t$.

Similarly, since the sum of $q, r, s, t$ and the sum of $r, s, t, u$ are both equal to 35 , and $r, s, t$ are added in each sum, then $q=u$.

It can similarly be shown that $r=v$ and $s=w$.

Using the above observations, the sequence $p, q, r, s, t, u, v, w$ can be written as $p, q, r, s, p, q, r, s$. Since the sum of $q$ and $v$ is 14 and $r=v$, then the sum of $q$ and $r$ is 14 .

Since the sum of $p, q, r, s$ is 35 and the sum of $q$ and $r$ is 14, then the sum of $p$ and $s$ is $35-14=21$.

The value of $p$ is as large as possible exactly when the value of $s$ is as small as possible.

Since $s$ is a positive integer, its smallest possible value is 1 .

Therefore, the largest possible value of $p$ is $21-1=20$.

(We note that 20,4,10,1,20,4,10,1 is an example of such a list.)

## Solution 2

The sum of the values of each group of four consecutive letters is 35 .

Thus, $p+q+r+s=35$ and $t+u+v+w=35$, and so $(p+q+r+s)+(t+u+v+w)=35+35=70$. Rearranging the sum of these eight letters, we get

$$
p+q+r+s+t+u+v+w=(p+w)+(q+v)+(r+s+t+u)=70
$$

However, $r+s+t+u=35$ (the sum of the values of four consecutive letters), and $q+v=14$. Substituting, we get $(p+w)+14+35=70$, and so $p+w=21$.

The value of $p$ is as large as possible exactly when the value of $w$ is as small as possible.

Since $w$ is a positive integer, its smallest possible value is 1 .

Substituting, we get $p+1=21$, and so the largest possible value of $p$ is 20 .

(We note that 20,12,2,1,20,12,2,1 is an example of such a list.)

ANSWER: (C)