AMC 10 · 2021 · #10
Grade 8 arithmeticProblem
Which of the following is equivalent to
(2+3)(22+32)(24+34)(28+38)(216+316)(232+332)(264+364)?
Pick an answer.
AMC 10 2021 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: Simplify the long product $(2 + 3)(2^2 + 3^2)(2^4 + 3^4)(2^8 + 3^8)(2^{16} + 3^{16})(2^{32} + 3^{32})(2^{64} + 3^{64})$. The exponents are $1, 2, 4, 8, 16, 32, 64$ — successive powers of $2$. Express the result in a clean closed form matching one of the choices.
Givens: $P = (2 + 3)(2^2 + 3^2)(2^4 + 3^4)(2^8 + 3^8)(2^{16} + 3^{16})(2^{32} + 3^{32})(2^{64} + 3^{64})$; $7$ factors total; the $k$-th factor is $(2^{2^{k-1}} + 3^{2^{k-1}})$ for $k = 1, 2, \ldots, 7$; Choices: (A) $3^{127} + 2^{127}$, (B) $3^{127} + 2^{127} + 2 \cdot 3^{63} + 3 \cdot 2^{63}$, (C) $3^{128} - 2^{128}$, (D) $3^{128} + 2^{128}$, (E) $5^{127}$
Unknowns: A closed-form simplification of $P$
Understand
Restated: Simplify the long product $(2 + 3)(2^2 + 3^2)(2^4 + 3^4)(2^8 + 3^8)(2^{16} + 3^{16})(2^{32} + 3^{32})(2^{64} + 3^{64})$. The exponents are $1, 2, 4, 8, 16, 32, 64$ — successive powers of $2$. Express the result in a clean closed form matching one of the choices.
Givens: $P = (2 + 3)(2^2 + 3^2)(2^4 + 3^4)(2^8 + 3^8)(2^{16} + 3^{16})(2^{32} + 3^{32})(2^{64} + 3^{64})$; $7$ factors total; the $k$-th factor is $(2^{2^{k-1}} + 3^{2^{k-1}})$ for $k = 1, 2, \ldots, 7$; Choices: (A) $3^{127} + 2^{127}$, (B) $3^{127} + 2^{127} + 2 \cdot 3^{63} + 3 \cdot 2^{63}$, (C) $3^{128} - 2^{128}$, (D) $3^{128} + 2^{128}$, (E) $5^{127}$
Plan
Primary tool: #5 Look for a Pattern
Secondary: #9 Solve an Easier Related Problem, #13 Convert to Algebra, #3 Eliminate Possibilities
Tool #9 (Easier Problem) first — try the product with $2$, then $3$ factors and watch what happens. Tool #5 (Pattern) reveals the doubling exponent in each collapse. Tool #13 (Algebra) supplies the engine, the difference of squares identity $(a - b)(a + b) = a^2 - b^2$. Tool #3 (Eliminate) discards the impostor choices by checking exponents.
Execute — Answer: C
6.EE.A.3 Step 1 - Try the easier version with just two factors.
- Multiply by $(3 - 2) = 1$ (free): $(3 - 2)(3 + 2)(3^2 + 2^2)$.
- Apply $(a-b)(a+b) = a^2 - b^2$ once: $(3^2 - 2^2)(3^2 + 2^2)$.
- Apply it again: $3^4 - 2^4$.
💡 Two factors give $3^4 - 2^4$ — the exponent doubled twice from $1$ to $4$, the count of original factors plus one.
5.OA.B.3 Step 2 - Pattern emerging.
- The exponent after collapsing $n$ factors is $2^n$, and the result is $3^{2^n} - 2^{2^n}$.
- Verify on $n = 3$ factors: $(3 + 2)(3^2 + 2^2)(3^4 + 2^4) \to 3^8 - 2^8$ after pre-multiplying by $(3 - 2)$ and chaining the identity three times.
💡 Each application of the difference of squares squares the previous exponent — so $1 \to 2 \to 4 \to 8 \to \ldots$
8.EE.A.1 Step 3 - Apply the pattern to the original product with $n = 7$ factors.
- Prepend $(3 - 2) = 1$, then collapse $8$ times in chain (one for the prepend pair, six more for the remaining factors, ending with one final squaring):
💡 Each collapse doubles the exponent: $1 \to 2 \to 4 \to 8 \to 16 \to 32 \to 64 \to 128$. Seven doublings produce $2^7 = 128$.
8.EE.A.1 Step 4 - Count carefully — there are $7$ original factors, so after prepending $(3-2)$ we have $8$ factors.
- Each application of $(a-b)(a+b) = a^2-b^2$ consumes one factor and doubles the exponent.
- After $7$ collapses (consuming $7$ of the $8$ factors), the running term is $3^{2^7} - 2^{2^7} = 3^{128} - 2^{128}$.
- The final factor has already been merged.
💡 $7$ original factors $\to$ $7$ doublings of the exponent starting from $1$ $\to$ end at $2^7 = 128$.
8.EE.A.1 Step 5 - Compare to the five choices.
- (A) $3^{127} + 2^{127}$: wrong exponent and wrong sign.
- (B) extra cross terms — wrong.
- (C) $3^{128} - 2^{128}$: matches.
- (D) $3^{128} + 2^{128}$: wrong sign — the identity produces a difference, not a sum.
- (E) $5^{127}$: the $5 = 3 + 2$ from the first factor is misleading — only the first factor equals $5$, the rest do not.
💡 The telescoping identity always produces $a^n - b^n$ (a difference), so any sum-form choice is impossible.
6.EE.A.3 Try the easier version with just two factors. Multiply by $(3 - 2) = 1$ (free): 5.OA.B.3 Pattern emerging. The exponent after collapsing $n$ factors is $2^n$, and the re 8.EE.A.1 Apply the pattern to the original product with $n = 7$ factors. Prepend $(3 - 2) 8.EE.A.1 Count carefully — there are $7$ original factors, so after prepending $(3-2)$ we 8.EE.A.1 Compare to the five choices. (A) $3^{127} + 2^{127}$: wrong exponent and wrong s Review
Reasonableness: Two independent sanity checks. First: count of factors $\to$ final exponent. $7$ factors with exponents $1, 2, 4, 8, 16, 32, 64$ — the sum $1+2+4+\ldots+64 = 127$ tempts choice (A), but the correct rule is doubling the highest exponent, giving $128$ — matches (C). Second: parity / sign — the difference-of-squares chain always produces $a^n - b^n$, never a sum, so (D) is wrong. (E) is wrong because $(3^2 + 2^2) = 13 \ne 5^2 = 25$.
Alternative: Tool #13 (Algebra) without the telescope: use the identity $(a^n - b^n) = (a - b)(a + b)(a^2 + b^2)(a^4 + b^4) \cdots (a^{n/2} + b^{n/2})$ for $n$ a power of $2$. With $n = 128$, the right side is exactly $(3 - 2)$ times the given product, so the product equals $(3^{128} - 2^{128}) / (3 - 2) = 3^{128} - 2^{128}$. Faster if the identity is already memorized.
CCSS standards used (min grade 8)
8.EE.A.1Know and apply the properties of integer exponents (Tracking the exponent through repeated squaring: $1 \to 2 \to 4 \to \ldots \to 128$ via $(a^k)^2 = a^{2k}$.)6.EE.A.3Apply the properties of operations to generate equivalent expressions (Applying $(a-b)(a+b) = a^2 - b^2$ to collapse pairs of factors.)5.OA.B.3Generate two numerical patterns using two given rules and identify relationships (Recognizing the doubling-exponent pattern from worked smaller cases ($n = 2, 3$ factors).)
⭐ This AMC 10 problem only needs Grade 8 exponent rules you already know — sneak in $(3 - 2) = 1$ at the front, then $(a - b)(a + b) = a^2 - b^2$ doubles the exponent each step. Seven doublings of $1$ land on $128$, giving $3^{128} - 2^{128}$.
⭐ This AMC 10 problem only needs Grade 8 exponent rules you already know — sneak in $(3 - 2) = 1$ at the front, then $(a - b)(a + b) = a^2 - b^2$ doubles the exponent each step. Seven doublings of $1$ land on $128$, giving $3^{128} - 2^{128}$.
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