1. In each part of this problem, seven different positive integers will be placed in the seven boxes of the "H"-shaped figure. The integers must be placed so that the three integers in the left vertical column, the three integers in the right vertical column, and the three integers in the one horizontal row all have the same sum. For example, in the "H" to the right, $8+11+12=31$ and $11+19+1=31$ and $27+1+3=31$.
(a) Place the integers 3, 5, 7, 15 in the figure so that each of the three sums is equal to 29 .
(b) There is a value of $t$ for which the figure shown has three equal sums and contains seven different integers. Determine this value of $t$.
(c) Seven different positive integers are placed in the figure so that the three sums are equal. If $a<c$, determine all possible values of $a$.
(d) The integers $k$ and $n$ are each between 4 and 18, inclusive. The figure contains seven different integers and the three sums are equal. Determine all possible values of $k$.
1. (a) Since the sum of the entries in the left column is 29 , the bottom left entry is $29-13-9=7$. Since the sum of the entries in the horizontal row is 29 , the remaining two entries have a sum of $29-9=20$. Since the possible entries are 3,5 and 15 , these entries must be 5 and 15 .
This means that the 3 must go in the bottom right box, which means that the middle box in the right column must contain $29-11-3=15$, which means that 5 goes in the middle box.
(b) The sums of the left and right columns are equal. Thus,
$$
\begin{aligned}
15+(t+1)+11 & =(2 t-3)+10+14 \\
t+27 & =2 t+21 \\
6 & =t
\end{aligned}
$$

If $t=6$, we complete the $\mathrm{H}$ to verify that the entries are all different:
(c) Since the three sums are equal, we have $a+12+c=12+9+7=b+7+11$ and so $a+c+12=28=b+18$ which gives $a+c=16$ and $b=10$.
Since $a+c=16$ and $a$ and $c$ are positive integers with $a<c$, we have the following possibilities:
- $a=1$ and $c=15$; all 7 integers are different in this case
- $a=2$ and $c=14$; all 7 integers are different in this case
- $a=3$ and $c=13$; all 7 integers are different in this case
- $a=4$ and $c=12$; this is not possible since there would be two $12 \mathrm{~s}$
- $a=5$ and $c=11$; this is not possible since there would be two 11 s
- $a=6$ and $c=10$; this is not possible since there would be two 10 s
- $a=7$ and $c=9$; this is not possible since there would be two 7s (and two 9s)
Thus, the possible values of $a$ are 1,2 and 3, giving the following figures:
(d) Since the three sums are equal, we have $7+10+k=10+n+18=(n+6)+18+4$ which gives $k+17=n+28=n+28$, which means that $k=n+11$.
Since $k$ is between 4 and 18 , inclusive, and $k=n+11$, then we must have $n \leq 7$. Since $n$ is between 4 and 18 , inclusive, then $n \geq 4$ as well.
Putting this together, $n$ can equal $4,5,6$, or 7 , which give values for $k$ of $15,16,17$, or 18 , respectively, and values of $n+6$ of $10,11,12$, or 13 , respectively.
Since the figure already contains 10 and 18 , we cannot have $k=18$ or $n+6=10$, which means that $n \neq 7$ and $n \neq 4$.
This means that the possible values of $k$ are 16 and 17 , which gives the following figures: