AMC 10 · 2016 · #20

Grade 8 algebra
function-evaluationfraction-arithmeticgcd logical-deductionconvert-to-algebra ↑ Prerequisites: function-evaluation
📏 Long solution 💡 4 insights
Problem
An unknown operation obeys two rules and one equation uses it twice. Report the solution's parts.

Pick an answer.

(A)
109
(B)
201
(C)
301
(D)
3049
(E)
33,601
How to solve
Strategy Introduce a Variable

The operation is never told to us directly, so treat the result a◆ b as an unknown expression and feed clever inputs into the two rules until the expression is forced. Substituting equal arguments collapses Rule 1 using Rule 2, which pins the operation down as ordinary division. After that the equation is routine: plug in, simplify, and solve a linear equation.

1STEP 1

Make two arguments equal

Making two arguments equal simplifies the first rule.

a◆(b◆ b)=(a◆ b) · b → a◆ 1=(a◆ b) · b
2STEP 2

Pin down a ◇ 1

That pins down one special value.

a◆ 1=(a◆ a) · a=1 · a=a
3STEP 3

Identify the operation

The operation is then recognisable.

(a◆ b) · b=a → a◆ b=a/b
4STEP 4

Check division fits both rules

It really does satisfy both rules.

a÷b/c=ac/b=a/b · c
5STEP 5

Plug into the equation

Substituting leaves a linear equation.

2016◆(6/x)=2016·x/6=336x=100
6STEP 6

Solve and reduce

Reducing gives 109, choice (A).

x=100/336=25/84, p+q=25+84=109
Answer
109
Check the chain forward with division: 6◆25/84=6÷25/84=504/25, then 2016÷504/25=2016·25/504=4·25=100, exactly the target. The reduced fraction 25/84 is between 0 and 1, a sensible nonzero real, and p+q=109 matches choice (A).
💡Key takeaway

When an operation is defined only by rules, feed it equal inputs to collapse the rules until you recognize a familiar operation hiding underneath.

  • Make two arguments equal
  • Pin down a ◇ 1
  • Identify the operation
  • Check division fits both rules
  • Plug into the equation
  • Solve and reduce