AMC 10 · 2002 · #3
Grade 6 arithmeticAccording to the standard convention for exponentiation,
2222=2(2(22))=216=65536.
If the order in which the exponentiations are performed is changed, how many other values are possible?
Pick an answer.
AMC 10 2002 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: A stack of four $2$'s, $2^{2^{2^{2}}}$, is worth $65536$ when the exponents are grouped in the standard right-to-left order. Insert the parentheses in every other possible order and count how many different values (not counting the given one) can come out.
Givens: The expression is a tower of exactly four $2$'s combined only by exponentiation; The standard grouping $2^{(2^{(2^{2})})}=2^{16}=65536$; Answer choices: (A) $0$, (B) $1$, (C) $2$, (D) $3$, (E) $4$
Unknowns: How many values other than $65536$ can be produced by re-parenthesizing the tower
Understand
Restated: A stack of four $2$'s, $2^{2^{2^{2}}}$, is worth $65536$ when the exponents are grouped in the standard right-to-left order. Insert the parentheses in every other possible order and count how many different values (not counting the given one) can come out.
Givens: The expression is a tower of exactly four $2$'s combined only by exponentiation; The standard grouping $2^{(2^{(2^{2})})}=2^{16}=65536$; Answer choices: (A) $0$, (B) $1$, (C) $2$, (D) $3$, (E) $4$
Plan
Primary tool: #2 Make a Systematic List
Secondary: #7 Identify Subproblems, #3 Eliminate Possibilities
"How many values are possible" is a counting question over a small, finite set of arrangements, so Tool #2 (Make a Systematic List) is the backbone: there are only a handful of ways to parenthesize a four-high tower, and writing every one of them down guarantees none is missed or double-counted. Tool #7 (Identify Subproblems) does the arithmetic safely — each grouping is evaluated one small power at a time from the innermost bracket outward, so no single step is bigger than squaring a two-digit number. Tool #3 (Eliminate Possibilities) finishes the job: after listing the results we strike out the duplicates and the original value, and what remains is the count the problem wants.
Execute — Answer: B
5.OA.A.1 Step 1 List every parenthesization
- A tower of four numbers joined by exponentiation can be grouped in exactly five ways.
- Writing them all out: $2^{\left(2^{\left(2^{2}\right)}\right)}$, $2^{\left(\left(2^{2}\right)^{2}\right)}$, $\left(2^{2}\right)^{\left(2^{2}\right)}$, $\left(\left(2^{2}\right)^{2}\right)^{2}$, and $\left(2^{\left(2^{2}\right)}\right)^{2}$.
- The first is the standard grouping from the problem; the goal is to see which of the other four give something new.
💡 Every different placement of parentheses is a different order of doing the exponent steps, so listing all five leaves nothing to chance.
6.EE.A.1 Step 2 Evaluate each from the inside out
- Work each grouping starting at the innermost bracket.
- Grouping 1: $2^{2}=4$, then $2^{4}=16$, then $2^{16}=65536$.
- Grouping 2: $2^{2}=4$, then $4^{2}=16$, then $2^{16}=65536$.
- Grouping 3: $2^{2}=4$ and $2^{2}=4$, then $4^{4}=256$.
- Grouping 4: $2^{2}=4$, then $4^{2}=16$, then $16^{2}=256$.
- Grouping 5: $2^{2}=4$, then $2^{4}=16$, then $16^{2}=256$.
- Notice the towers only ever collapse to two sizes.
💡 Squaring a modest number stays small, but stacking a $16$ up in the exponent is what blows the tower up to $65536$.
5.OA.A.1 Step 3 Discard the repeats and the original
- The five results are $65536,\,65536,\,256,\,256,\,256$, so only two distinct values ever appear: $65536$ and $256$.
- The value $65536$ is the one already given in the problem, which the question tells us to exclude.
- That leaves a single new value, $256$.
- So exactly one other value is possible, which is choice (B).
💡 Counting the OTHER values means throwing away both the duplicates and the value the problem started with.
5.OA.A.1 A tower of four numbers joined by exponentiation can be grouped in exactly five 6.EE.A.1 Work each grouping starting at the innermost bracket. Grouping 1: $2^{2}=4$, the 5.OA.A.1 The five results are $65536,\,65536,\,256,\,256,\,256$, so only two distinct val Review
Reasonableness: There are only five ways to parenthesize the tower and one of them is the given expression, so at most $4$ other values could ever exist — the count must be between $0$ and $4$, matching the answer choices. It cannot be $0$, because $256\neq 65536$ shows a genuinely different value does appear. Finding exactly one new value is believable: an exponent tower of $2$'s collapses to just two sizes, the huge $2^{16}=65536$ when a $16$ lands up in the exponent, and the modest $256$ whenever the final step is a squaring. Two sizes, minus the original, leaves one.
Alternative: Argue by the outermost step instead of listing all five. The last operation is either $2^{(\text{tower of three})}$ or $(\text{tower of three})^{2}$. Every three-high tower of $2$'s equals $16$, so the first shape always gives $2^{16}=65536$ and the second always gives $16^{2}=256$. Only these two shapes exist, so besides the given $65536$ there is exactly one other value, confirming (B).
CCSS standards used (min grade 6)
5.OA.A.1Use parentheses, brackets, or braces in numerical expressions and evaluate (Listing the five ways to parenthesize the tower and understanding that the parenthesis placement is exactly what changes the value, then discarding duplicates and the original.)6.EE.A.1Write and evaluate numerical expressions involving whole-number exponents (Evaluating each grouping one power at a time from the innermost bracket outward: $2^{2}=4$, $4^{2}=16$, $2^{4}=16$, $16^{2}=256$, and $2^{16}=65536$.)
⭐ Parentheses decide the order of the exponent steps, and a four-high tower of 2's only ever collapses to two sizes — 65536 or 256 — so beyond the value given there is just one other.
⭐ Parentheses decide the order of the exponent steps, and a four-high tower of 2's only ever collapses to two sizes — 65536 or 256 — so beyond the value given there is just one other.
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