Question

For any set $S,$ let $|S|$ denote the number of elements in $S,$ and let $n(S)$ be the number of subsets of $S,$ including the empty set and the set $S$ itself. If $A,$ $B,$ and $C$ are sets for which $\begin{gathered} n(A)+n(B)+n(C)\\ {}=n(A\cup B\cup C),\\ |A|=|B|=100, \end{gathered}$ then what is the minimum possible value of $|A\cap B\cap C|?$

$96$

$97$

$98$

$99$

$100$

Answer: B

Concepts:set cardinality subsets powers of two extremal counting

Difficulty rating: 2430

Small Hint:

Use $n(S)=2^{|S|}$ and determine $|C|$ and $|A\cup B\cup C|$

Big Hint:

Within the union, count how many elements each of $A,$ $B,$ and $C$ can omit

Solution:

Let $c=|C|$ and $u=|A\cup B\cup C|.$ Since $n(S)=2^{|S|},$ the equation becomes $2^{101}+2^c=2^u.$ The two summands must be equal, so $c=101$ and $u=102.$ Within the union, $A,$ $B,$ and $C$ omit $2,$ $2,$ and $1$ elements. Thus at most five distinct elements are absent from the triple intersection, giving $|A\cap B\cap C|\ge97.$ Equality is attained by making those omissions distinct.

Thus the correct answer is B.

1991 AMC 12 Problem 30

Answer: B

Difficulty rating: 2430

Small Hint:

Big Hint:

Solution:

Problem 30 in Other Years