AMC 10 · 2016 · #23
Grade 8 algebraPick an answer.
AMC 10 2016 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
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.
Make two arguments equal
Put c=b in Rule 1; by Rule 2 the inner b◇b=1, so it collapses to a◇1=(a◇b) · b.
Feeding equal inputs lets Rule 2 erase the messy inner term and expose a simpler relationship.
6.EE.B.6Introduce A VariablePin down a ◇ 1
Now set b=a; the right side is (a◇a) · a and a◇a=1, so a◇1=a.
A second smart substitution turns one unknown into a number we actually know.
6.EE.A.2Introduce A VariableIdentify the operation
Combine a◇1=(a◇b) · b with a◇1=a: then (a◇b) · b=a, so a◇b= — the operation is division.
Once a quantity times b equals a, that quantity has to be a divided by b.
Once a quantity times one number gives another, that quantity has to be the second divided by the first.
▸ Why?
Dividing undoes multiplying, so it recovers the factor exactly.
▸ Why?
That factor times its partner returns one, which is what makes the recovery exact.
Check division fits both rules
Test a◇b=: Rule 2 gives =1, and Rule 1 gives a÷== · c. Both hold.
A candidate rule is only safe once it survives every condition it must satisfy.
6.EE.A.2Guess And CheckPlug into the equation
Swap ◇ for division: 6◇x=, then 2016÷=2016 · =336x, so 336x=100.
Dividing by a fraction flips it, so the x climbs into the numerator and the problem becomes linear.
6.NS.A.1Introduce A VariableSolve and reduce
From 336x=100, x== in lowest terms, so p=25, q=84 and p+q=109. Answer (A).
Isolate x, then strip the common factor so p and q are truly relatively prime.
8.EE.C.7Work BackwardsWhen 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