AMC 10 Step by Step

2006 AMC 10A Problem 2

P1-10~1 minPrint

Define x⊗y=x3−yx\otimes y=x^3-y . What is h⊗(h⊗h)h\otimes (h\otimes h) ?

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Key insight

Do the inside operation first. Then use its result as the second input of the outside operation, keeping it inside parentheses.

Solution

The rule x⊗y=x3−yx \otimes y = x^3 - y means: cube the first number, then subtract the second number.

1. Start with the inside part, because the outside part needs its value. With both inputs equal to hh:
h⊗h=h3−h.h \otimes h = h^3 - h.
2. Now the outside operation has first input hh and second input h3−hh^3 - h. Cube the first input and subtract the whole second input, in parentheses:
h⊗(h3−h)=h3−(h3−h).h \otimes (h^3 - h) = h^3 - (h^3 - h).
3. Remove the parentheses. The minus sign flips both signs inside:
h3−h3+h=h.h^3 - h^3 + h = h.

The answer is (C) h\boxed{\textbf{(C)}\ h}.

Why this works

A made-up operation is just a rule with two slots. Here the two slots are treated differently (only the first one gets cubed), so it matters which number goes where. Always work from the innermost parentheses outward, and when you plug in an expression, keep it in parentheses so the minus sign reaches every term.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

Work from the inside out. The rule says "cube the first input and subtract the second," so
h⊗h=h3−h.h \otimes h = h^3 - h.
Now the outer operation has first input hh and second input h3−hh^3 - h:
h⊗(h3−h)=h3−(h3−h)=h3−h3+h=h.h \otimes (h^3 - h) = h^3 - (h^3 - h) = h^3 - h^3 + h = h.

The answer is (C) h\boxed{\textbf{(C)}\ h}.

Review in the AoPS books

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Problem © Mathematical Association of America (MAA), American Mathematics Competitions. Reproduced for non-commercial educational use. Solution and commentary are original to this site. Also appeared as 2006 12A #2.