A7. Carry has six cards. On each card, a positive integer is written. She chooses three cards and calculates the sum of the numbers on those cards. She does this for all 20 possible combinations of three cards. Ten times Carry gets a result of 16 and ten times a result of 18. What is the smallest number that appears on the cards?
A) 1
B) 2
C) 3
D) 4
E) 5
A7. D) 4 Because there are only two different outcomes, only two different numbers can appear on the cards. Indeed, if three cards bear different numbers, then in combination with a pair of the remaining three cards, they would each yield a different result. Let the two numbers that appear be $a$ and $b$. We can assume that $a$ appears on at least three cards. Since $a+a+a$, $a+a+b$, and $a+b+b$ are all different, $b$ can only appear once.

There are now two cases: $a+a+a=16$, $a+a+b=18$ and $a+a+a=18$, $a+a+b=16$. The first possibility is ruled out because 16 is not a multiple of three (since $a$ is an integer). Therefore, we see that $a=\frac{18}{3}=6$ and $b=16-12=4$. The smallest number that appears is thus 4.