Question

Let $a\mathbin{@}b$ represent the operation on two numbers, $a$ and $b,$ which selects the larger of the two numbers, with $a\mathbin{@}a=a.$ Let $a\mathbin{!}b$ represent the operation which selects the smaller of the two numbers, with $a\mathbin{!}a=a.$ Which of the following three rules is (are) correct? $\begin{aligned} (1)\quad&a\mathbin{@}b=b\mathbin{@}a,\\ (2)\quad&a\mathbin{@}(b\mathbin{@}c)=(a\mathbin{@}b)\mathbin{@}c,\\ (3)\quad&a\mathbin{!}(b\mathbin{@}c)\\ &\quad=(a\mathbin{!}b)\mathbin{@}(a\mathbin{!}c). \end{aligned}$

$(1)$ only

$(2)$ only

$(1)$ and $(2)$ only

$(1)$ and $(3)$ only

all three

Answer: E

Concepts:custom operation distributive property casework

Difficulty rating: 1500

Small Hint:

Translate $\mathbin{@}$ as maximum and $\mathbin{!}$ as minimum

Big Hint:

For rule $(3),$ compare both sides separately when $a$ is below or above $\max(b,c)$

Solution:

Maximum is commutative and associative, so $(1)$ and $(2)$ hold. Rule $(3)$ is the distributive identity $\begin{aligned} &\min(a,\max(b,c))\\ &\quad=\max(\min(a,b),\min(a,c)). \end{aligned}$ If $a\ge\max(b,c),$ both sides equal $\max(b,c);$ if $a\lt\max(b,c),$ both sides equal $a.$ Thus $(3)$ also holds.

Therefore, the correct answer is E.

1962 AMC 12 Problem 27

Answer: E

Difficulty rating: 1500

Small Hint:

Big Hint:

Solution:

Problem 27 in Other Years