AMC 10 · 2012 · #8
Grade 7 arithmeticWhat is the sum of all integer solutions to 1<(x−2)2<25?
Pick an answer.
AMC 10 2012 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 every integer $x$ that makes the compound inequality $1<(x-2)^2<25$ true, then add all of those integer values together.
Givens: The compound inequality $1<(x-2)^2<25$ must hold; $x$ must be an integer; Both inequality signs are strict, so the endpoints $1$ and $25$ are not allowed; Answer choices: (A) $10$, (B) $12$, (C) $15$, (D) $19$, (E) $25$
Unknowns: The sum of all integer values of $x$ that satisfy $1<(x-2)^2<25$
Understand
Restated: Find every integer $x$ that makes the compound inequality $1<(x-2)^2<25$ true, then add all of those integer values together.
Givens: The compound inequality $1<(x-2)^2<25$ must hold; $x$ must be an integer; Both inequality signs are strict, so the endpoints $1$ and $25$ are not allowed; Answer choices: (A) $10$, (B) $12$, (C) $15$, (D) $19$, (E) $25$
Plan
Primary tool: #4 Introduce a Variable
Secondary: #3 Eliminate Possibilities, #2 Make a Systematic List
The expression $(x-2)$ repeats, so Tool #4 (Introduce a Variable) renames it as $u=x-2$ and turns the problem into the clean inequality $1<u^2<25$. Tool #3 (Eliminate Possibilities) is exactly what the strict $<$ signs demand: throw out the boundary cases where $u^2$ equals $1$ or $25$. Tool #2 (Make a Systematic List) then lays out every surviving solution so none is missed before adding.
Execute — Answer: B
6.EE.B.6 Step 1 Rename the repeated chunk
- The whole inequality is built around $(x-2)$, so let $u=x-2$.
- It becomes $1<u^2<25$, which is easier to reason about because it involves just one squared quantity.
💡 Naming the repeated part with a single letter turns a cluttered expression into a simple one about $u$.
6.EE.B.5 Step 2 Find which integers survive
- I need integers $u$ whose square is strictly between $1$ and $25$.
- Testing squares: $2^2=4$, $3^2=9$, and $4^2=16$ all fall between $1$ and $25$, but $1^2=1$ and $5^2=25$ hit the forbidden endpoints and are out.
- A negative squares to the same value, so both signs count.
- Thus $u\in\{-4,-3,-2,2,3,4\}$.
💡 The strict signs quietly delete the boundary cases $u=\pm1$ and $u=\pm5$.
6.NS.C.5 Step 3 Translate back to $x$
- Undo the rename with $x=u+2$.
- Adding $2$ to each $u$ gives $x\in\{-2,-1,0,4,5,6\}$.
- Listing every value keeps all six solutions in view so none is missed or double-counted.
💡 Each $u$ maps to exactly one $x$, so six $u$-values give six $x$-values, and some come out negative.
7.NS.A.1 Step 4 Add all the solutions
- Sum the six integer solutions: $-2+(-1)+0+4+5+6$.
- The negatives $-2$ and $-1$ cancel part of the positive total, leaving $12$.
- So the answer is (B).
💡 Adding signed values just means the negatives and positives partly cancel before you total up.
6.EE.B.6 The whole inequality is built around $(x-2)$, so let $u=x-2$. It becomes $1<u^2< 6.EE.B.5 I need integers $u$ whose square is strictly between $1$ and $25$. Testing squar 6.NS.C.5 Undo the rename with $x=u+2$. Adding $2$ to each $u$ gives $x\in{-2,-1,0,4,5,6\ 7.NS.A.1 Sum the six integer solutions: $-2+(-1)+0+4+5+6$. The negatives $-2$ and $-1$ ca Review
Reasonableness: Use symmetry as a check: the six $u$-values $\pm2,\pm3,\pm4$ are balanced around $0$, so they add to $0$. Since $x=u+2$, the six $x$-values add to $0+6\times2=12$, matching the direct sum. The result $12$ is choice (B). The classic trap is forgetting the strict inequalities and letting $u=\pm1$ or $u=\pm5$ slip in, which would change both the count and the total.
Alternative: Skip the substitution and read the bound as a distance: $1<(x-2)^2<25$ means $|x-2|$ is strictly between $1$ and $5$, so $|x-2|\in\{2,3,4\}$. That gives $x-2=\pm2,\pm3,\pm4$ and the same six values $x\in\{-2,-1,0,4,5,6\}$, which sum to $12$.
CCSS standards used (min grade 7)
6.EE.B.6Use variables to represent numbers and write expressions to solve problems (Introducing $u=x-2$ to rewrite the inequality as $1<u^2<25$.)6.EE.A.1Write and evaluate numerical expressions involving whole-number exponents (Evaluating the squares $2^2,3^2,4^2$ (and the boundary $1^2,5^2$) to test the inequality.)6.EE.B.5Understand solving an equation or inequality as a process of finding values (Deciding which integer values of $u$ actually make $1<u^2<25$ true, excluding the strict boundaries.)6.NS.C.5Understand that positive and negative numbers describe quantities (Recognizing that both signs of $u$ work and that some resulting $x$-values are negative.)7.NS.A.1Apply and extend understanding of addition and subtraction to rational numbers (Adding the signed solutions $-2+(-1)+0+4+5+6=12$.)
⭐ Rename the repeated chunk with a letter, use the strict $<$ signs to throw out the boundary values, then list and add every solution.
⭐ Rename the repeated chunk with a letter, use the strict $<$ signs to throw out the boundary values, then list and add every solution.
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