1962 AMC 12 Problem 27

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

Translate

\mathbin{@}

as maximum and

\mathbin{!}

as minimum

For rule

(3),

compare both sides separately when

a

is below or above

\max(b,c)

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.

Problem 27 in other years