Skip to main content

2016 AMC 10A Problem 23

Problem 23 of 25HarderAlgebra

A binary operation \diamondsuit has the properties that a(bc)=(ab)ca\,\diamondsuit\, (b\,\diamondsuit \,c) = (a\,\diamondsuit \,b)\cdot c and that aa=1a\,\diamondsuit \,a=1 for all nonzero real numbers a,a, b,b, and c.c. (Here the dot \cdot represents the usual multiplication operation.) The solution to the equation 2016(6x)=1002016 \,\diamondsuit\, (6\,\diamondsuit\, x)=100 can be written as pq,\frac{p}{q}, where pp and qq are relatively prime positive integers. What is p+q?p+q?

Answer choices

Show solution

Solution

Since aa=1a\diamondsuit a=1, substituting b=cb=c in a(bc)=(ab)ca\diamondsuit(b\diamondsuit c)=(a\diamondsuit b)c gives a1=(ab)ba\diamondsuit1=(a\diamondsuit b)b. Also, using a(aa)=(aa)aa\diamondsuit(a\diamondsuit a)=(a\diamondsuit a)a gives a1=aa\diamondsuit1=a. Therefore ab=aba\diamondsuit b=\frac ab. The equation becomes 2016(6x)=20166x=20166x=336x=100. \begin{aligned} 2016\diamondsuit(6\diamondsuit x) &= 2016\diamondsuit\frac6x \\ &= \frac{2016}{\frac{6}{x}} \\ &= 336x \\ &= 100. \end{aligned} Thus x=2584x=\frac{25}{84}, so p+q=25+84=109p+q=25+84=109. Thus, the correct answer is A.

More practice

Concepts: custom operation · functional equation

Problem text and solution from the LIVE past-contest archive. See also the AoPS wiki page for community solutions.