AMC 10 Step by Step
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2010 AMC 10B Problem 1

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

What is 100(100−3)−(100⋅100−3)100(100-3)-(100\cdot100-3) ?

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Key insight

Multiply out 100(100 - 3) to get 10000 - 300. The 10000 shows up in both parts and cancels, leaving -300 + 3.

Solution

Both parts contain 100⋅100=10000100 \cdot 100 = 10000, so instead of multiplying everything out, look for that piece to cancel.

1. First part: multiply the 100100 into the parentheses.
100(100−3)=100⋅100−100⋅3=10000−300.100(100 - 3) = 100 \cdot 100 - 100 \cdot 3 = 10000 - 300.
2. Second part: 100⋅100−3=10000−3100 \cdot 100 - 3 = 10000 - 3.
3. Now subtract. A minus sign in front of parentheses changes the sign of everything inside, so
−(10000−3)=−10000+3.-(10000 - 3) = -10000 + 3.
4. Put the pieces together:
10000−300−10000+3.10000 - 300 - 10000 + 3.
The 1000010000 and the −10000-10000 cancel, leaving −300+3=−297-300 + 3 = -297.

The answer is (C) −297\boxed{\textbf{(C)}\ -297}.

Why this works

When two parts of an expression share a big piece, write both parts so the shared piece is visible and let it cancel. The one thing to be careful about is the minus sign in front of the second parentheses: it flips the −3-3 into +3+3.

Alternative approach

Just do the arithmetic. 100⋅97=9700100 \cdot 97 = 9700 and 10000−3=999710000 - 3 = 9997, so the answer is 9700−9997=−2979700 - 9997 = -297.

Concise solution

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

Expand the first product with the distributive property:
100(100−3)=100⋅100−300.100(100-3) = 100\cdot 100 - 300.
The second parenthesis is 100⋅100−3100\cdot 100 - 3. Subtracting it flips both of its signs:
(100⋅100−300)−(100⋅100−3)=−300+3=−297.(100\cdot 100 - 300) - (100\cdot 100 - 3) = -300 + 3 = -297.
The 100⋅100100\cdot 100 terms cancel, and no large multiplication is ever needed.

The answer is (C) −297\boxed{\textbf{(C)}\ -297}.

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.