, 2013

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 $(n+3)+(n+4)=2n+7$; the sum of every other pair is smaller. The sum of any two of these integers is at least $n+(n+1)=2n+1$; the sum of every other pair is larger. Therefore, the maximum possible difference between the sums of two pairs is $(2n+7)-(2n+1)$ 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 $2n+7=63$ or $2n=56$ 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 $n+(n+1)+(n+2)+(n+3)+(n+4)=5n+10$. 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 $5n+10=120+r$ or $r=5n-110$. Since $r=5n-110=5(n-22)$, 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, $r=30$.