AMC 10 · 2024 · #7
Grade 6 number-theoryProblem
What is the remainder when is divided by ?
Pick an answer.
Toolkit + CCSS Solution
Understand
Restated: Find the remainder when the sum $7^{2024} + 7^{2025} + 7^{2026}$ is divided by $19$.
Givens: Three consecutive powers of $7$: exponents $2024$, $2025$, $2026$; The sum of the three powers is divided by $19$; Answer choices: (A) $0$, (B) $1$, (C) $7$, (D) $11$, (E) $18$
Unknowns: The remainder $r$ with $0 \le r < 19$
Understand
Restated: Find the remainder when the sum $7^{2024} + 7^{2025} + 7^{2026}$ is divided by $19$.
Givens: Three consecutive powers of $7$: exponents $2024$, $2025$, $2026$; The sum of the three powers is divided by $19$; Answer choices: (A) $0$, (B) $1$, (C) $7$, (D) $11$, (E) $18$
Plan
Primary tool: #13 Convert to Algebra
Secondary: #7 Identify Subproblems
The three terms all share $7^{2024}$, so the algebra move is to factor it out and inspect the bracket $1 + 7 + 7^2$. That is Tool #13 (Convert to Algebra) in its cleanest form — a numerical sum becomes a product whose individual factors can be checked for divisibility one at a time. Tool #7 (Identify Subproblems) then breaks the divisibility question into two pieces: compute $1 + 7 + 49 = 57$, then check whether $57$ is divisible by $19$. If yes, the whole sum is divisible by $19$ regardless of what $7^{2024}$ equals, and the remainder is $0$.
Execute — Answer: A
6.EE.A.3 Step 1 - Factor the smallest power $7^{2024}$ out of every term.
- Using $7^{2025} = 7^{2024} \cdot 7$ and $7^{2026} = 7^{2024} \cdot 7^2$, the sum becomes a single product.
💡 Pulling out a common factor turns three messy terms into one — Grade 6 "apply the distributive property to generate equivalent expressions."
6.EE.A.1 Step 2 - Subproblem A: evaluate the bracket.
- The arithmetic inside the parentheses is small enough to do by hand.
💡 Just plug in $7^2 = 49$ and add — Grade 6 evaluating a numerical expression with whole-number exponents.
4.OA.B.4 Step 3 - Subproblem B: check whether $57$ is divisible by $19$.
- Try $19 \times 3$ — that's exactly $57$, so $57 = 3 \times 19$.
💡 Recognizing $57 = 3 \times 19$ is Grade 4 "find factor pairs / determine multiples" — the only fact this whole problem really depends on.
6.NS.B.4 Step 4 - Substitute the factorization of $57$ back.
- Now $19$ appears as a factor of the whole expression, so the sum is a multiple of $19$ and the remainder is $0$.
💡 A number with $19$ as a factor leaves remainder $0$ on division by $19$ — the basic meaning of "multiple of" from Grade 6.
6.EE.A.3 Factor the smallest power $7^{2024}$ out of every term. Using $7^{2025} = 7^{202 6.EE.A.1 Subproblem A: evaluate the bracket. The arithmetic inside the parentheses is sma 4.OA.B.4 Subproblem B: check whether $57$ is divisible by $19$. Try $19 \times 3$ — that' 6.NS.B.4 Substitute the factorization of $57$ back. Now $19$ appears as a factor of the w Review
Reasonableness: Cross-check with modular arithmetic: $7^2 = 49 = 2 \times 19 + 11$, so $7^2 \equiv 11 \pmod{19}$. Then $7^3 \equiv 7 \cdot 11 = 77 = 4 \times 19 + 1$, so $7^3 \equiv 1 \pmod{19}$. Three consecutive powers $7^{2024}, 7^{2025}, 7^{2026}$ cycle through values $\{7^a, 7^a \cdot 7, 7^a \cdot 7^2\}$ which sum to $7^a(1 + 7 + 49) \equiv 7^a \cdot 57 \equiv 7^a \cdot 0 = 0 \pmod{19}$. Same answer (A).
Alternative: Tool #5 (Look for a Pattern): compute $1 + 7 + 7^2$ modulo $19$ first ($= 57 \equiv 0$), then notice that the same bracket factor appears no matter what exponent base you start from. So $7^k + 7^{k+1} + 7^{k+2}$ is always a multiple of $19$ for every $k \ge 0$ — the constant-$0$ pattern across all $k$ confirms answer (A).
CCSS standards used (min grade 6)
6.EE.A.3Apply the properties of operations to generate equivalent expressions (Factoring the common power $7^{2024}$ out of all three terms to rewrite the sum as $7^{2024}(1 + 7 + 7^2)$.)6.EE.A.1Write and evaluate numerical expressions involving whole-number exponents (Evaluating the bracket $1 + 7 + 7^2 = 1 + 7 + 49 = 57$.)4.OA.B.4Find all factor pairs and recognize multiples; determine prime or composite (Recognizing $57 = 3 \times 19$ as a factor pair, which makes $19$ a factor of the whole sum.)6.NS.B.4Find greatest common factor and least common multiple of two numbers (Concluding that any number with $19$ as a factor leaves remainder $0$ on division by $19$ — the divisibility meaning of "multiple of $19$.")
⭐ This AMC 10 problem only needs Grade 6 factoring — pull the common power out, check whether $19$ divides the small leftover, and you don't need to compute any huge power at all!
⭐ This AMC 10 problem only needs Grade 6 factoring — pull the common power out, check whether $19$ divides the small leftover, and you don't need to compute any huge power at all!