AMC 10 Step by Step

2016 AMC 10B Problem 2

P1-10~1 minPrint

If n♡m=n3m2n\heartsuit m=n^3m^2 , what is 2♡44♡2\frac{2\heartsuit 4}{4\heartsuit 2} ?

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Key insight

Follow the rule exactly: cube the first number and square the second. Then 2 heart 4 is 128 and 4 heart 2 is 256.

Solution

The rule says: cube the first number, square the second number, and multiply.

1. For 2♡42 \heartsuit 4, the first number is 22 and the second is 44:
2♡4=23⋅42=8⋅16=128.2 \heartsuit 4 = 2^3 \cdot 4^2 = 8 \cdot 16 = 128.
2. For 4♡24 \heartsuit 2, the first number is 44 and the second is 22:
4♡2=43⋅22=64⋅4=256.4 \heartsuit 2 = 4^3 \cdot 2^2 = 64 \cdot 4 = 256.
3. Divide:
128256=12.\frac{128}{256} = \frac12.

The answer is (B) 12\boxed{\textbf{(B)}\ \frac{1}{2}}.

Why this works

A made-up symbol like ♡\heartsuit is just a recipe. The only skill is putting each number into the right place in the recipe. Writing out both values in full before dividing keeps you from swapping the roles by accident.

Alternative approach

Use letters to see a shortcut. The fraction is
23⋅4243⋅22.\frac{2^3 \cdot 4^2}{4^3 \cdot 2^2}.
Cancel 222^2 from 232^3 (leaving one 22) and cancel 424^2 from 434^3 (leaving one 44 on the bottom). What is left is 24=12\frac{2}{4} = \frac12.

Concise solution

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

The first input is cubed and the second is squared.

2♡4=23⋅42=8⋅16=128,4♡2=43⋅22=64⋅4=256.2\heartsuit 4 = 2^3 \cdot 4^2 = 8 \cdot 16 = 128, \qquad 4\heartsuit 2 = 4^3 \cdot 2^2 = 64 \cdot 4 = 256.

Therefore
2♡44♡2=128256=12.\frac{2\heartsuit 4}{4\heartsuit 2} = \frac{128}{256} = \frac12.

The answer is (B) 12\boxed{\textbf{(B)}\ \frac{1}{2}}.

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.