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2000 AMC 10 Problem 2

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2000 AMC 10 Problem 2AMC 10 Step by Step

Name: Date: September 13, 2026

Exponents, Logarithms & Radicals · difficulty 1 of 5 · about 1 min · position P1-10

$2000(2000^{2000}) = x$ Find x.

A)
$2000^{2001}$
B)
$4000^{2000}$
C)
$2000^{4000}$
D)
$4,000,000^{2000}$
E)
$2000^{4,000,000}$

Answer

Key insight

Multiplying by one more factor of 2000 raises the exponent by one: 2000^1 times 2000^2000 is 2000^2001.

Solution

Write the leading $2000$ as $2000^1$ . Both factors now share the base $2000$ , and powers of the same base multiply by adding exponents:
$2000^{1} \cdot 2000^{2000} = 2000^{1 + 2000} = 2000^{2001}.$

The answer is $\boxed{\textbf{(A)}\ 2000^{2001}}$ .

Why this works

$2000^{2000}$ is a string of $2000$ copies of $2000$ multiplied together; tacking on one more copy makes $2001$ copies. The exponent counts copies, so it goes up by exactly one. Bases never multiply together unless the exponents match, and exponents never multiply unless a power is raised to a power.

Alternative approach

Sanity-check the rule on tiny numbers: $2 \cdot 2^3 = 16 = 2^4$ , whereas $4^3 = 64$ and $2^6 = 64$ are wrong. The same pattern picks (A).

The trap

Multiplying the bases to get 4000^2000, or multiplying the exponents to get 2000^4000.

Common mistakes

Multiplying the bases to get 4000^2000, or multiplying the exponents to get 2000^4000.
Choosing (E) by confusing $a \cdot a^n$ with $(a^n)^n$ ; exponents multiply only when a power is raised to another power.

Techniques

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