2016 AMC 10A Problem 23

Question

A binary operation

\diamondsuit

has the properties that

a\,\diamondsuit\, (b\,\diamondsuit \,c) = (a\,\diamondsuit \,b)\cdot c

and that

a\,\diamondsuit \,a=1

for all nonzero real numbers

a,

b,

and

c.

(Here the dot

\cdot

represents the usual multiplication operation.) The solution to the equation

2016 \,\diamondsuit\, (6\,\diamondsuit\, x)=100

can be written as

\frac{p}{q},

where

p

and

q

are relatively prime positive integers. What is

p+q?

109

201

301

3049

33{,}601

First prove

a\diamondsuit b=\frac{a}{b}

Then compute

2016\diamondsuit(6\diamondsuit x)

Since

a\diamondsuit a=1

, substituting

b=c

in

a\diamondsuit(b\diamondsuit c)=(a\diamondsuit b)c

gives

a\diamondsuit1=(a\diamondsuit b)b

. Also, using

a\diamondsuit(a\diamondsuit a)=(a\diamondsuit a)a

gives

a\diamondsuit1=a

. Therefore

a\diamondsuit b=\frac ab

.

The equation becomes

\begin{aligned} 2016\diamondsuit(6\diamondsuit x) &= 2016\diamondsuit\frac6x \\ &= \frac{2016}{\frac{6}{x}} \\ &= 336x \\ &= 100. \end{aligned}

Thus

x=\frac{25}{84}

, so

p+q=25+84=109

.

Thus, the correct answer is A.

Problem 23 in other years