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2016 AMC 10A Problem 1

P1-10~1 minPrint
Algebraic ManipulationSet up the equation/formula and compute; no special trick needed

What is the value of 11!−10!9!\dfrac{11!-10!}{9!} ?

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Key insight

11! is just 11 times 10!, so the top is 10 copies of 10!. Then 10! divided by 9! is 10, giving 10 times 10.

Solution

Never multiply out 11!11! by hand. Instead, notice how neighboring factorials are linked.

1. Recall that 11!=11⋅10⋅9⋯111! = 11 \cdot 10 \cdot 9 \cdots 1, which is the same as 11⋅10!11 \cdot 10!. In the same way, 10!=10⋅9!10! = 10 \cdot 9!.
2. So the top of the fraction is
11!−10!=11⋅10!−1⋅10!.11! - 10! = 11 \cdot 10! - 1 \cdot 10!.
That is "1111 copies of 10!10! take away 11 copy of 10!10!", which leaves 1010 copies:
11!−10!=10⋅10!.11! - 10! = 10 \cdot 10!.
3. Now divide by 9!9!. Since 10!=10⋅9!10! = 10 \cdot 9!, dividing 10!10! by 9!9! gives 1010:
10⋅10!9!=10⋅10=100.\frac{10 \cdot 10!}{9!} = 10 \cdot 10 = 100.

The answer is (B) 100\boxed{\textbf{(B)}\ 100}.

Why this works

Each factorial is the one before it times one more number, so neighboring factorials share almost everything. Pull out the common factorial first and only small numbers are left to multiply. Whenever you see factorials added, subtracted or divided, look for the largest factorial they all contain.

Alternative approach

Split the fraction into two pieces and simplify each one:
11!9!−10!9!=(11⋅10)−10=110−10=100.\frac{11!}{9!} - \frac{10!}{9!} = (11 \cdot 10) - 10 = 110 - 10 = 100.

Concise solution

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

Both terms of the numerator contain 10!10!, since 11!=11⋅10!11! = 11 \cdot 10!. Factor it out:
11!−10!=11⋅10!−10!=10⋅10!.11! - 10! = 11 \cdot 10! - 10! = 10 \cdot 10!.

Now divide by 9!9!. Because 10!=10⋅9!10! = 10 \cdot 9!, we get
10⋅10!9!=10⋅10=100.\frac{10 \cdot 10!}{9!} = 10 \cdot 10 = 100.

The answer is (B) 100\boxed{\textbf{(B)}\ 100}.

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.