The four positive integers $a, b, c, d$ satisfy $a<b<c<d$. When the sums of the six pairs of these integers are calculated, the six answers are all different and the four smallest sums are 6 , 8,12 , and 21 . What is the value of $d$ ?
The six pairs are $a+b, a+c, a+d, b+c, b+d, c+d$.

The sum of the two smallest numbers will be the smallest sum. Thus, $a+b=6$.

The second smallest sum will be $a+c$ since $a<b<c<d$ gives $a+b<a+c$ and $a+c<a+d$ and $a+c<b+c$ and $a+c<b+d$ and $a+c<c+d$. Thus, $a+c=8$.

Using a similar argument, the largest sum must be $c+d$ and the second largest sum must be $b+d$.

Therefore, the third and fourth sums (as listed in increasing order) are $a+d=12$ and $b+c=21$ or $b+c=12$ and $a+d=21$.

If $b+c=21$, then we would have $(a+b)+(a+c)+(b+c)=6+8+21$ and so $2 a+2 b+2 c=35$. This cannot be the case since $a, b, c$ are integers, which makes the left side even and the right side odd.

Therefore, $b+c=12$ and $a+d=21$.

This gives $(a+b)+(a+c)+(b+c)=6+8+12$ or $2 a+2 b+2 c=26$ or $a+b+c=13$.

Since $b+c=12$, then $a=(a+b+c)-(b+c)=13-12=1$.

Since $a+d=21$, then $d=21-a=20$.

(We can check that $a=1, b=5, c=7, d=20$ fit with the given information.)

ANSWER: 20