In the diagram, $p, q, r, s$, and $t$ represent five consecutive integers, not necessarily in order. The two integers in the leftmost circle add to 63 . The two integers in the rightmost circle add to 57 . What is the value of $r$ ?
(A) 24
(B) 28
(C) 20
(D) 42
(E) 30

![](https://cdn.mathpix.com/cropped/2024_04_20_ac36362783317e0251fdg-064.jpg?height=198&width=398&top_left_y=579&top_left_x=1297)
Solution 1

Suppose that the five consecutive integers represented by $p, q, r, s, t$ are $n, n+1, n+2, n+3, n+4$, for some integer $n$.

The sum of any two of these integers is at most $(n+3)+(n+4)=2 n+7$; the sum of every other pair is smaller.

The sum of any two of these integers is at least $n+(n+1)=2 n+1$; the sum of every other pair is larger.

Therefore, the maximum possible difference between the sums of two pairs is $(2 n+7)-(2 n+1)$ or 6 ; any other choice of pairs will give a smaller difference between the sums.

Since we are told that $p+q=63$ and $s+t=57$, which gives $(p+q)-(s+t)=6$, then it must be the case that $p$ and $q$ are the two largest integers from the list while $s$ and $t$ are the
two smallest integers from the list.

In other words, $p+q=(n+3)+(n+4)=63$ and so $2 n+7=63$ or $2 n=56$ and so $n=28$.

Since $r$ must be the middle integer in the list, then $r=n+2=30$.

## Solution 2

Suppose that the five consecutive integers represented by $p, q, r, s, t$ are $n, n+1, n+2, n+3, n+4$, for some integer $n$.

The sum of all five integers is $n+(n+1)+(n+2)+(n+3)+(n+4)=5 n+10$.

We are told that $p+q=63$ and $s+t=57$.

Thus, the sum of the five integers is also $p+q+r+s+t=63+r+57=120+r$.

Comparing the two expressions for the sum of the integers, we obtain $5 n+10=120+r$ or $r=5 n-110$.

Since $r=5 n-110=5(n-22)$, then $r$ is divisible by 5 .

Of the five given answer choices, this means that we could have $r=20$ or $r=30$.

If $r=20$, then $20=5(n-22)$ or $n-22=4$ and so $n=26$. In this case, $r$ is not one of the integers between $n$ and $n+4$, inclusive, so $r$ cannot be 20 .

If $r=30$, then $30=5(n-22)$ or $n-22=6$ and so $n=28$. Here, the integers in the list would be $28,29,30,31,32$, which can produce the given conditions if $p$ and $q$ are 31 and 32 , and $t$ and $s$ are 28 and 29 .

Therefore, $r=30$.

ANSWER: (E)