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2010 AMC 10B Problem 1

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2010 AMC 10B Problem 1AMC 10 Step by Step
Name:                             Date: September 13, 2026
Algebraic Manipulation · difficulty 1 of 5 · about 1 min · position P1-10

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

  1. A)

    −20,000-20,000

  2. B)

    −10,000-10,000

  3. C)

    −297-297

  4. D)

    −6-6

  5. E)

    00

Answer

C

Key insight

Distribute the first product: 100(100-3) = 100*100 - 300, so the two 100*100 terms cancel and only -300 + 3 remains.

Solution

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}.

Why this works

When two expressions share a big common piece, expand just enough to see the shared piece and let it cancel. Subtracting a parenthesized quantity negates every term inside it, which is exactly where the sign of the −3-3 matters.

Alternative approach

Compute both sides numerically: 100⋅97=9700100\cdot 97 = 9700 and 10000−3=999710000 - 3 = 9997, so the difference is 9700−9997=−2979700 - 9997 = -297.

The trap

Dropping the parentheses and computing 100*100 - 3 - 100*100 - 3 = -6, choice (D).

Common mistakes

  • Dropping the parentheses and computing 100⋅100−3−100⋅100−3=−6100\cdot100 - 3 - 100\cdot100 - 3 = -6, choice (D).
  • Forgetting to distribute the 100100 over the −3-3 and using 100⋅100−3100\cdot100 - 3 for the first term, which gives 00.

Techniques

Set up the equation/formula and compute; no special trick needed