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2002 AMC 10A Problem 3

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2002 AMC 10A Problem 3AMC 10 Step by Step
Name:                             Date: September 13, 2026
Exponents, Logarithms & Radicals · difficulty 2 of 5 · about 3 min · position P1-10

According to the standard convention for exponentiation, 2222=2(2(22))=216=65536.2^{2^{2^{2}}} = 2^{(2^{(2^2)})} = 2^{16} = 65536.

If the order in which the exponentiations are performed is changed, how many other values are possible?

  1. A)

    00

  2. B)

    11

  3. C)

    22

  4. D)

    33

  5. E)

    44

Answer

B

Key insight

There are only five ways to parenthesize four 2s; since 2^(2^2) = (2^2)^2 = 16, every grouping evaluates to either 2^16 or 256.

Solution

A chain of four numbers with three operations can be fully parenthesized in exactly five ways. Write xyx^y as x∧yx \wedge y to keep the groupings readable:

  1. 2∧(2∧(2∧2))2 \wedge (2 \wedge (2 \wedge 2)): the standard order, 216=655362^{16} = 65536.
  2. 2∧((2∧2)∧2)2 \wedge ((2 \wedge 2) \wedge 2): inner part is 42=164^2 = 16, so again 2162^{16}.
  3. ((2∧2)∧2)∧2((2 \wedge 2) \wedge 2) \wedge 2: (42)2=162=256(4^2)^2 = 16^2 = 256.
  4. (2∧(2∧2))∧2(2 \wedge (2 \wedge 2)) \wedge 2: (24)2=162=256(2^4)^2 = 16^2 = 256.
  5. (2∧2)∧(2∧2)(2 \wedge 2) \wedge (2 \wedge 2): 44=2564^4 = 256.

The key coincidence is 2(22)=(22)2=162^{(2^2)} = (2^2)^2 = 16, which is why groupings 1 and 2 agree and groupings 3 and 4 agree. Every non-standard grouping gives 256256, so exactly one value other than 6553665536 occurs.

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

Why this works

Exponentiation is not associative, so different groupings can genuinely differ, but with such a short chain a complete list is the fastest proof. The useful facts are (ab)c=abc(a^b)^c = a^{bc} and the special coincidence 24=422^4 = 4^2; together they collapse five expressions into two values.

The trap

Counting parenthesizations instead of distinct values, or forgetting that the problem asks for values other than the standard one.

Common mistakes

  • Counting parenthesizations instead of distinct values, or forgetting that the problem asks for values other than the standard one.
  • Evaluating (22)(22)(2^2)^{(2^2)} as 2(22)⋅22^{(2^2)\cdot 2} or some other misuse of the power rule; the exponents multiply only in the form (ab)c=abc(a^b)^c = a^{bc}.

Techniques

Organized listing / direct enumeration