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2002 AMC 10A Problem 3

P1-10~3 minPrint
Exponents, Logarithms & RadicalsBasic CountingOrganized listing / direct enumeration

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?

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Key insight

There are only five ways to group four 2s. Because 2 to the 4th and 4 squared are both 16, the five groupings give just two different values.

Solution

There are only a few ways to group four 22s, so the safest plan is to list all of them and work each one out.

  1. How many groupings are there? Think about which power is done last. It splits the four 22s into a left group and a right group: 11 and 33, or 22 and 22, or 33 and 11. A group of three 22s can itself be grouped 22 ways, and a group of one or two 22s only 11 way. So there are 2+1+2=52 + 1 + 2 = 5 groupings.
  1. Work out each one, always starting with the innermost parentheses:
groupingstepsvalue
2(2(22))2^{(2^{(2^2)})}22=42^2 = 4, 24=162^4 = 16, 2162^{16}6553665536
2((22)2)2^{((2^2)^2)}22=42^2 = 4, 42=164^2 = 16, 2162^{16}6553665536
((22)2)2((2^2)^2)^222=42^2 = 4, 42=164^2 = 16, 16216^2256256
(2(22))2(2^{(2^2)})^222=42^2 = 4, 24=162^4 = 16, 16216^2256256
(22)(22)(2^2)^{(2^2)}22=42^2 = 4, 22=42^2 = 4, 444^4256256
  1. The first row is the standard order. Why do so many rows agree? Because 242^4 and 424^2 are both 1616, so two different-looking middle steps land on the same number.
  1. The only values that appear are 6553665536 and 256256. Apart from the standard value 6553665536, exactly one other value, 256256, is possible.

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

Why this works

Changing the order of powers can change the answer, but when there are only a handful of orders, listing them all is the quickest and most reliable proof. The lucky fact 24=42=162^4 = 4^2 = 16 is what squeezes five groupings down to two values. Read the question carefully: it asks for other values, not other groupings.

Concise solution

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

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}.

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. Also appeared as 2002 12A #3.