A merchant received four bags of potatoes and wants to measure the weight of each one. He knows that the weights of these bags in kilograms are whole and distinct numbers. Suppose the weights of the bags (in kilograms) are $a, b, c$, and $d$, with $a<b<c<d$.

a) Show that, when weighing the bags two at a time, the largest result is $c+d$, and the second largest is $b+d$. Also show that the smallest result is $a+b$ and the second smallest is $a+c$.

b) The merchant's scale broke, so it can only indicate weights greater than or equal to 100 kilograms. He decides to weigh the bags two at a time, performing six weighings. In four of the weighings, he obtained the following measurements (in kilograms): 101, 112, 116, and 127. In the other two, he only found out that the sums were less than or equal to 100 kilograms. Find the weights of the four bags.
Solution

a) Adding $a$ to each side of the inequality $b<c$, we obtain the inequality $a+b<a+c$. With similar arguments, we can obtain the inequalities:

$$
a+c<a+d, \quad a+c<b+c, \quad a+d<b+d, \quad b+c<b+d \quad \text { and } \quad b+d<c+d
$$

Using these inequalities, we can conclude that:

$$
a+b<a+c<a+d<b+d<c+d \quad \text { and } \quad a+b<a+c<b+c<b+d<c+d
$$

From (.15), we see that the 4 sums $a+b, a+c, a+d$ and $b+c$ are less than the sums $b+d \mathbf{and} c+d$. Since $b+d<c+d$, the largest sum is $c+d$ and the second largest is $b+d$. Also from (.15), we see that $a+b$ and $a+c$ are less than each of the other 4 sums $a+d, b+c, b+d$ and $c+d$. Since $a+b<a+c$, then $a+b$ is the smallest sum and $a+c$ is the second smallest.

b) From part a) we know that the largest sum is $c+d$ and that the second largest is $b+d$. Thus we have

$$
c+d=127 \quad \text { and } \quad b+d=116
$$

Therefore, the sums $a+d$ and $b+c$ are equal to the numbers 101 and 112 in some order. Subtracting the equations in (.16), we get $c=b+11$. Adding $b$ to each side, we see that $b+c=2b+11$. Since $2b+11$ is an odd number, we discover that $b+c$ is odd. Thus,

$$
b+c=101 \quad \text { and } \quad a+d=112
$$

Using that $c=b+11 \mathrm{in}$ (.17) we get $b=45$. Using that $b=45 \mathrm{in}$ (.16) and (.17), we see that $c=56$ and $d=71$. Finally, we get that $a=41$.