A Pre-RMO question! -16

As $a>b>c>d \therefore a+b>a+c;b+c>b+d>c+d$ now either $a+d>b+c$ or $b+c>a+d$

Case $1$ ( $a+d>b+c$ ) :

As $x+y$ have to be largest $\therefore x$ and $y$ have to be the largest among $200,185,211,222$ . Assuming $x>y$ $a+b=x.....[1]$ $a+c=y.....[2]$ $a+d=222.....[3]$ $b+c=211.....[4]$ $b+d=200.....[5]$ $c+d=185.....[6]$

$[4]-[5] \Rightarrow c-d=11.....[7]$ $[3]-[6] \Rightarrow a-c=37.....[8]$ $[7]+[8] \Rightarrow a-d=48.....[9]$ $[9]+[3] \Rightarrow \boxed{2a=270}$ $[1]+[2] \Rightarrow x+y=2a+b+c=270+211=481$ $4+8+1=\boxed{13}$

Case $2$ ( $b+c>a+d$ ) :

As $x+y$ have to be largest $\therefore x$ and $y$ have to be the largest among $200,185,211,222$ . Assuming $x>y$ $a+b=x.....[1]$ $a+c=y.....[2]$ $b+c=222.....[3]$ $a+d=211.....[4]$ $b+d=200.....[5]$ $c+d=185.....[6]$

$[3]-[5] \Rightarrow c-d=22.....[7]$ $[4]-[6] \Rightarrow a-c=26.....[8]$ $[7]+[8] \Rightarrow a-d=48.....[9]$ $[9]+[4] \Rightarrow \boxed{2a=259}$ $[1]+[2] \Rightarrow x+y=2a+b+c=259+222=481$ $4+8+1=\boxed{13}$

First let us compare sum of all possible pairs of distinct elements of the set $\{a\,,b\,,c\,,d\}$ such that $a > b > c > d$ .

Since $a > b$ and $b > c$ . Adding both the equation,

$\Rightarrow a + b > b + c \hspace{20pt}\cdots\text{Eq. 1}$

Since $b > c$ and $c > d$ . Adding both the equation,

$\Rightarrow b + c > c + d \hspace{20pt}\cdots\text{Eq. 2}$

Since $b > c$

$\Rightarrow a + b > a + c \hspace{20pt}\cdots\text{Eq. 3}$

Since $a > b$

$\Rightarrow a + c > b + c \hspace{20pt}\cdots\text{Eq. 4}$

Since $c > d$

$\Rightarrow b + c > b + d \hspace{20pt}\cdots\text{Eq. 5}$

Since $b > c$

$\Rightarrow b + d > c + d \hspace{20pt}\cdots\text{Eq. 6}$

From Eq. 1, Eq. 2, Eq. 3, Eq. 4, Eq. 5 and Eq. 6 we get relation between 5 pairwise sum:

$a + b > a + c > b + c > b + d > c + d$

Since $c > d$

$\Rightarrow a + c > a + d \hspace{20pt}\cdots\text{Eq. 7}$

Since $a > b$

$\Rightarrow a + d > b + d \hspace{20pt}\cdots\text{Eq. 8}$

From Eq. 4, Eq. 5, Eq. 7 and Eq. 8, two cases arises:

Case 1 : $a + c > a + d > b + c > b + d$

Case 2 : $a + c > b + c > a + d > b + d$

So, overall two cases occurs:

Case 1 : $a + b > a + c > a + d > b + c > b + d > c + d$

Given pairwise sums are $185, 200, 211, 222, x$ and $y$ . To get the maximum possible possible values of $x$ and $y$ , we take $x = a + b$ and $y = a + c$ . The only possible allocations of value are:

$c + d = 185\newline b + d = 200 \newline b + c = 211 \newline a + d = 222$

Solving these $4$ equations for $a\,,b\,,c$ and $d$ we get,

$a = 135\,,\,b = 113\,,\,c = 98\,,\,d = 87$

So, $x = a + b = 135 + 113 = 248$ and $y = a + c = 135 + 98 = 233$

$\Rightarrow x + y = 248 + 233 = 481$

Case 2 : $a + b > a + c > b + c > a + d > b + d > c + d$

Agian, to get maximum possible value $x$ and $y$ we take $x = a + b$ and $y = a + c$ . The only possible allocations of value are:

$c + d = 185\newline b + d = 200 \newline a + d = 211 \newline b + c = 222$

Solving these $4$ equations for $a\,,b\,,c$ and $d$ we get,

$a = 129.5\,,\,b = 118.5\,,\,c = 103.5\,,\,d = 81.5$

So, $x = a + b = 129.5 + 118.5 = 248$ and $y = a + c = 129.5 + 103.5 = 233$

$\Rightarrow x + y = 248 + 233 = 481$

We see that from both the cases, value of $x + y$ is same.

So, maximum possible value of $x + y$ is $481$ .

Hence sum of digits of $x + y = 4 + 8 + 1 = \boxed{13}$ .

Let $(m,n)$ be $m+n$ and $x > y$

• greatest possible value of $x+y$ means x is the largest and y is the second largest.
• clearly, $(a,b)= x$ - is the largest and $(c,d)=185$ - is the smallest
• As $(a,c) > (b,c) , (a,d) > (b,d),$ we compare $(b,c) > (b,d)$ so $(b,d) = 200$
• In the remaining 3 pairwise sums, $(a,c) >$ both $(b,c)$ and $(a,d)$ . so $(a,c) = y$
• $\textcolor{#D61F06}{(a,d) , (b,c)}$ will match to either $222$ or $211$ .

$x = (a,b) \\ y = (a,c) \\ 222 = ? \\ 211 = ? \\ 200= (b,d) \\ 185 = (c,d)$

• $\textcolor{#D61F06}{(a,b,c,d) = 222 + 211 = 433 !}$
• $(a,b,c,d) - (b,d) = (a,c) = 433 - 200 = 233 = y$
• $(a,b,c,d) - (c,d) = (a,b) = 433 - 185 = 248 = x$
• $x + y = 233+248 = 481$

Therefore, the digits of the greatest possible value of $x+y = 4+8+1 = \boxed{13}$

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if we look for integers solution then

• $(b,d) - (c,d) = b-c = 15 \implies b=c+15$

• $(b, c) = (c+15, c)$ which is odd, $\therefore (b,c) = 211$ and $(a,d) =222$

• $2c+15=211 \implies c = 98$

• $b=98+15= 113$
• $d = 185-98 = 87$
• $a=433-113-98-87=135$
• ${a,b,c,d} = {135,113,98,87}$

$x + y = 2a + b + c = 2(135)+113 +98 = 481$