This question is really simple if you focus on this part
“The sum of any two is divisible by the third”
<span>⟹⟹</span> Pairwise, the numbers must have a GCD greater than one.
Let this GCD be <span>xx</span>.
The first three easiest numbers I can think of are <span><span>x,2x</span><span>x,2x</span></span> and <span><span>3x</span><span>3x</span></span>
Because
<span><span>gcd(x,2x)=gcd(2x,3x)=gcd(3x,x)=x</span><span>gcd(x,2x)=gcd(2x,3x)=gcd(3x,x)=x</span></span>
The numbers sum to <span>9696</span>
<span><span>⟹x+2x+3x=96</span><span>⟹x+2x+3x=96</span></span>
<span><span>⟹6x=96</span><span>⟹6x=96</span></span>
<span><span>⟹x=16</span><span>⟹x=16</span></span>
The largest number is <span><span>3x=3×16=48</span><span>3x=3×16=48</span></span>
Thanks for the A2A
