AMC 8 · 2016 · #15
Grade 6 number-theoryalgebraProblem
What is the largest power of 2 that is a divisor of 134−114?
Pick an answer.
AMC 8 2016 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: Find the biggest power of $2$ (that is, $2^k$ for the largest possible $k$) that divides the number $13^4 - 11^4$ exactly, with no remainder.
Givens: The target expression is $13^4 - 11^4$; $13^4 = 28561$ and $11^4 = 14641$ (so the difference is $13920$, but we will avoid computing it directly); Answer choices: (A) $8$, (B) $16$, (C) $32$, (D) $64$, (E) $128$
Unknowns: The largest power of $2$ that divides $13^4 - 11^4$
Understand
Restated: Find the biggest power of $2$ (that is, $2^k$ for the largest possible $k$) that divides the number $13^4 - 11^4$ exactly, with no remainder.
Givens: The target expression is $13^4 - 11^4$; $13^4 = 28561$ and $11^4 = 14641$ (so the difference is $13920$, but we will avoid computing it directly); Answer choices: (A) $8$, (B) $16$, (C) $32$, (D) $64$, (E) $128$
Plan
Primary tool: #7 Identify Subproblems
Secondary: #9 Solve an Easier Related Problem
Computing $13^4 - 11^4 = 13920$ directly is doable but messy, and we then still have to factor $13920$. Tool #7 (Identify Subproblems) gives a cleaner path: use the difference-of-squares identity $a^2 - b^2 = (a-b)(a+b)$ twice to break $13^4 - 11^4$ into three small factors — $(13-11)$, $(13+11)$, and $(13^2 + 11^2)$ — and count the $2$s in each factor separately. Tool #9 (Easier Related Problem) backs this up: testing $a^2 - b^2 = (a-b)(a+b)$ on a tiny case like $5^2 - 3^2$ makes the identity obvious, so the factoring move is trustworthy. We avoid Tool #13 (Algebra) and brute-force division because the factored form does the work for us.
Execute — Answer: C
6.EE.A.1 Step 1 - Confirm the difference-of-squares identity on an easier case before using it on $13$ and $11$.
- Try $5^2 - 3^2$: directly, $25 - 9 = 16$; via the identity, $(5-3)(5+3) = 2 \times 8 = 16$.
- They match, so the rule $a^2 - b^2 = (a-b)(a+b)$ is safe to use.
💡 Verifying an identity on small numbers first is the Tool #9 "easier problem" move — it builds confidence before applying the rule to harder numbers.
6.EE.A.1 Step 2 Treat $13^4$ as $(13^2)^2$ and $11^4$ as $(11^2)^2$, then apply the identity once to split the fourth-power difference into two factors.
💡 Splitting one hard expression into two easier ones is the core Tool #7 "subproblems" move.
6.EE.A.1 Step 3 - The first factor $13^2 - 11^2$ is itself a difference of squares, so apply the identity again to split it further.
- This gives three simple factors in place of the original fourth-power difference.
💡 Re-applying the same easier-problem split to a leftover piece is exactly how subproblems compound — each call peels off a layer.
5.NBT.B.5 Step 4 - Compute the third factor as a plain whole-number sum.
- $13^2 = 169$ and $11^2 = 121$, so $13^2 + 11^2 = 290$.
- The full product to analyze is $2 \times 24 \times 290$.
💡 Each factor is now a number we can hold in our head, so the hard expression has shrunk to three friendly pieces.
6.NS.B.4 Step 5 - Count the $2$s in each of the three factors using prime factorization.
- Then add the exponents — when factors multiply, their powers of $2$ add.
- $2 = 2^1$.
- $24 = 8 \times 3 = 2^3 \times 3$.
- $290 = 2 \times 145$, and $145 = 5 \times 29$ is odd, so $290 = 2^1 \times 145$.
- Adding exponents: $1 + 3 + 1 = 5$, so the largest power of $2$ dividing the product is $2^5 = 32$.
💡 Prime factorization tells you each factor's "$2$-content"; the exponent rule $2^a \times 2^b = 2^{a+b}$ lets you total them in one step.
6.EE.A.1 Confirm the difference-of-squares identity on an easier case before using it on 6.EE.A.1 Treat $13^4$ as $(13^2)^2$ and $11^4$ as $(11^2)^2$, then apply the identity onc 6.EE.A.1 The first factor $13^2 - 11^2$ is itself a difference of squares, so apply the i 5.NBT.B.5 Compute the third factor as a plain whole-number sum. $13^2 = 169$ and $11^2 = 1 6.NS.B.4 Count the $2$s in each of the three factors using prime factorization. Then add Review
Reasonableness: Direct check: $13^4 = 28561$, $11^4 = 14641$, so $13^4 - 11^4 = 13920$. Halve repeatedly: $13920 \to 6960 \to 3480 \to 1740 \to 870 \to 435$. We divided by $2$ exactly $5$ times before reaching an odd number, so $13920 = 2^5 \times 435 = 32 \times 435$. That matches the factored answer $32$ exactly, so choice (C) is correct. Choice (D) $64$ would have needed $6$ halvings, and choice (B) $16$ would have stopped after only $4$.
Alternative: Tool #6 (Guess and Check) on the answer choices: compute $13^4 - 11^4 = 13920$, then test divisibility by each option from biggest to smallest. $13920 \div 128 = 108.75$ (not whole, eliminate E). $13920 \div 64 = 217.5$ (eliminate D). $13920 \div 32 = 435$ (whole — candidate C). Since $32$ works and $64$ does not, the largest valid power of $2$ is $32$, confirming (C).
CCSS standards used (min grade 6)
5.NBT.B.5Fluently multiply multi-digit whole numbers (Computing $13^2 = 169$, $11^2 = 121$, and the sum $169 + 121 = 290$ to evaluate the third factor.)6.EE.A.1Write and evaluate numerical expressions involving whole-number exponents (Rewriting $13^4 - 11^4$ as $(13^2)^2 - (11^2)^2$ and applying the difference-of-squares identity at the level of exponential expressions.)6.NS.B.4Find the greatest common factor and use prime factorization (Prime-factoring each of $2$, $24$, and $290$ to count the powers of $2$ and combining them via $2^a \times 2^b = 2^{a+b}$.)
⭐ This AMC 8 problem only needs the Grade 6 idea of prime factorization plus one factoring trick — $a^2 - b^2 = (a-b)(a+b)$ — that you can verify with tiny numbers like $5^2 - 3^2$!
⭐ This AMC 8 problem only needs the Grade 6 idea of prime factorization plus one factoring trick — $a^2 - b^2 = (a-b)(a+b)$ — that you can verify with tiny numbers like $5^2 - 3^2$!
More like this
Same archetype — closest grade level first.