25. Christina has eight coins whose weights in grams are different positive integers. When Christina puts any two coins in one pan of her balance scales and any two in the other pan of the balance scales, the side containing the heaviest of those four coins is always the heavier side.
What is the smallest possible weight of the heaviest of the eight coins?
A 8
B 12
C 34
D 55
E 256
Solution
$\mathbf{C}$

Let the coins in such a set have weights (in grams) $a, b, c, d, e, f, g, h$ in ascending order of weight. Note that, to check a set has the stated property, it is enough to check, for each coin, that it together with the lightest coin is heavier than the pair of coins immediately preceding it. Suppose such a set has the minimal value for $h$. Then $a=1$ because, if $a \geq 2$ then a matching set in which each coin's weight has been reduced by 1 would still have the property and have a smaller value for $h$.
Also $b \geq 2$ and $c \geq 3$ because the weights are distinct integers. Applying the check to $d$, we get $d+1>b+c \geq 2+3=5$. Hence $d \geq 5$. Similarly, $e+1>c+d \geq 3+5=8$, hence $e \geq 8$. Continuing in this way we end up with $h \geq 34$. Moreover, the argument shows that the set of weights $1,2,3,5,8,13,21,34$ does have the required property. Hence the minimum value for the heaviest coin is 34 grams.