AMC 10 · 2012 · #8
Grade 7 arithmeticPick an answer.
AMC 10 2012 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
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 ＜ u² ＜ 25. Tool #3 (Eliminate Possibilities) is exactly what the strict ＜ signs demand: throw out the boundary cases where u² equals 1 or 25. Tool #2 (Make a Systematic List) then lays out every surviving solution so none is missed before adding.
Rename the repeated chunk
The chunk (x-2) repeats, so let u=x-2. The inequality becomes 1 ＜ u² ＜ 25 — one squared quantity instead of a cluttered one.
Naming the repeated part with a single letter turns a cluttered expression into a simple one about u.
6.EE.B.6Introduce A VariableFind which integers survive
Squares strictly inside 1 and 25 are 4, 9, 16; the endpoints are out. Both signs count, so u∈{-4,-3,-2,2,3,4}.
The strict signs quietly delete the boundary cases u=±1 and u=±5.
6.EE.B.5Eliminate PossibilitiesTranslate back to x
Undo the rename with x=u+2: adding 2 to each u gives x∈{-2,-1,0,4,5,6}, all six solutions in one list.
Each u maps to exactly one x, so six u-values give six x-values, and some come out negative.
Each value of the renamed chunk maps back to exactly one value of the original unknown.
▸ Why?
The renaming pairs the two sets off one for one, so no solution is gained or lost.
▸ Why?
Undoing the substitution reverses the step exactly, which is what recovers each original value.
Add all the solutions
Add them: -2+(-1)+0+4+5+6. The negatives cancel part of the positives, leaving 12 — choice (B).
Adding signed values just means the negatives and positives partly cancel before you total up.
7.NS.A.1Make A Systematic ListRename 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
- Find which integers survive
- Translate back to x
- Add all the solutions