The 4 Numbers

The option list is not needed.

Since the chosen numbers are distinct, say they are $a<b<c<d$ . From the information given, $d=a+7$ and $c=b+2$ . Since $a+b+c+d=31$ , we have

$a+b+b+2+a+7=2(a+b)+9=31$

so $b=11-a$ . This means the value of $a$ uniquely determines the list of numbers as $a,11-a,13-a,a+7$ . From $a<b<c<d$ , we have $a<11-a$ , so $a<6$ ; also $13-a<a+7$ , so $a>3$ .

Finally, note that $a$ has the opposite parity to all of the other numbers. We're told that only one of the numbers is odd; so $a$ is odd. The only possibility we're left with is $a=5$ , giving the answer $\boxed{5,6,8,12}$ .

Since the largest number minus the smallest number is odd, one of them must be odd. If the largest number be odd, then there is no solution. Therefore the smallest number must be odd. Let the numbers be 2q+1, 2m, 2n, 2p. Then 2(m+n+p+q)=31-1=30, or m+n+p+q=15. Let 2m be the largest number. Then 2m-2q-1=7 or m=q+4. Also 2p-2n=2. This implies p=n+1. Then (q+4)+(n+1)+q+n=15, or q+n=5. There two integer solutions to this equation: q=1, n=4 and q=2, n=3. The first yields m=p=5, which contradicts the condition that all the numbers are distinct. The second yields m=6, p=4. Then the numbers are (2)(2)+1=5, (2)(3)=6, (2)(4)=8 and (2)(6)=12