21. A grocer has twelve weights, weighing $1,2,3,4,5,6,7,8,9,10,11,12$ kilograms respectively. He splits them into three groups of four weights each. The total weights of the first and second groups are $41 \mathrm{~kg}$ and $26 \mathrm{~kg}$ respectively. Which of the following weights is in the same group as the $9 \mathrm{~kg}$ weight?
A $3 \mathrm{~kg}$
B $5 \mathrm{~kg}$
C $7 \mathrm{~kg}$
D $8 \mathrm{~kg}$
E $10 \mathrm{~kg}$
SolUTION
$\mathbf{C}$

The total weight is $1+2+\ldots+12 \mathrm{~kg}=78 \mathrm{~kg}$. The first two groups weigh $41 \mathrm{~kg}$ and $26 \mathrm{~kg}$, leaving $(78-41-26) \mathrm{kg}=11 \mathrm{~kg}$ for the third group. Since this is only $1 \mathrm{~kg}$ heavier than the smallest possible combination of $(1+2+3+4) \mathrm{kg}=10 \mathrm{~kg}$, there is only one way to combine 4 weights to get $11 \mathrm{~kg}$, namely $(1+2+3+5) \mathrm{kg}$. The next smallest combination would then be $(4+6+7+8) \mathrm{kg}=25 \mathrm{~kg}$, so the only way to get $26 \mathrm{~kg}$ would be $(4+6+7+9) \mathrm{kg}=26 \mathrm{~kg}$. Hence the $9 \mathrm{~kg}$ weight is in the same group as the $7 \mathrm{~kg}$.