AMC 10 · 2009 · #7
Grade 5 algebraBy inserting parentheses, it is possible to give the expression
2×3+4×5
several values. How many different values can be obtained?
Pick an answer.
AMC 10 2009 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: The expression $2\times3+4\times5$ has three operations. By adding parentheses you may force some operations to happen before others, which can change the result. Count how many different numbers the expression can equal over all legal ways of inserting parentheses.
Givens: The fixed number sequence is $2,\ 3,\ 4,\ 5$ with operations $\times,\ +,\ \times$ between them, in that order; Parentheses may be inserted, but the numbers and the operation signs keep their order; Answer choices: (A) $2$, (B) $3$, (C) $4$, (D) $5$, (E) $6$
Unknowns: The number of distinct values the parenthesized expression can produce
Understand
Restated: The expression $2\times3+4\times5$ has three operations. By adding parentheses you may force some operations to happen before others, which can change the result. Count how many different numbers the expression can equal over all legal ways of inserting parentheses.
Givens: The fixed number sequence is $2,\ 3,\ 4,\ 5$ with operations $\times,\ +,\ \times$ between them, in that order; Parentheses may be inserted, but the numbers and the operation signs keep their order; Answer choices: (A) $2$, (B) $3$, (C) $4$, (D) $5$, (E) $6$
Plan
Primary tool: #2 Make a Systematic List
Secondary: #7 Identify Subproblems, #3 Eliminate Possibilities
The question is 'how many different values', so the safe route is to list every legal grouping and evaluate it (Tool #2). Four numbers in a row have exactly five full parenthesizations, a small enough set to write out completely with no guessing. Each grouping is then a short arithmetic subproblem to evaluate (Tool #7). Finally, because two groupings can land on the same number, Tool #3 (Eliminate Possibilities) is used to drop the duplicates so only distinct values remain.
Execute — Answer: C
5.OA.A.1 Step 1 See what parentheses control
- There are three operations in $2\times3+4\times5$, and parentheses decide the order they run in.
- Four numbers in a row can be fully parenthesized in exactly five ways, so there are five expressions to check — no more, no fewer.
💡 Parentheses don't move the numbers; they only choose which operation goes first, so listing the groupings lists every possible outcome.
5.OA.A.2 Step 2 Write out the five groupings
- List all five full parenthesizations of $2\times3+4\times5$: group from the left in stages, put the plus in the middle, or group from the right.
- Writing each one down makes sure none is missed or repeated.
💡 Recording each grouping as its own expression turns a fuzzy 'try parentheses' into a fixed checklist of five things to evaluate.
5.OA.A.1 Step 3 Evaluate each grouping
- Work each expression from the innermost parentheses outward.
- In order: $\big((2\times3)+4\big)\times5=(6+4)\times5=50$; $\big(2\times(3+4)\big)\times5=(2\times7)\times5=70$; $(2\times3)+(4\times5)=6+20=26$; $2\times\big((3+4)\times5\big)=2\times35=70$; $2\times\big(3+(4\times5)\big)=2\times23=46$.
💡 Doing the inside of the parentheses first is exactly what the parentheses are telling you to do.
5.OA.A.1 Step 4 Count the distinct values
- Collect the five results $\{50,70,26,70,46\}$ and drop the repeat: the second and fourth groupings both give $70$.
- The different values that survive are $26,\ 46,\ 50,\ 70$ — four of them.
- So the expression can take $4$ different values, which is choice (C).
💡 'Different values' means each number counts once, so equal results must be merged before counting.
5.OA.A.1 There are three operations in $2\times3+4\times5$, and parentheses decide the or 5.OA.A.2 List all five full parenthesizations of $2\times3+4\times5$: group from the left 5.OA.A.1 Work each expression from the innermost parentheses outward. In order: $\big((2\ 5.OA.A.1 Collect the five results $\{50,70,26,70,46\}$ and drop the repeat: the second an Review
Reasonableness: The two multiplications $2\times3$ and $4\times5$ can each be forced or delayed by grouping, and the smallest possible result comes from doing both multiplications first, $26$, while the largest comes from folding the $+$ inside so everything gets multiplied, $70$. The four surviving values $26,46,50,70$ all sit inside that range and are clearly different, so $4$ is believable and matches choice (C).
Alternative: Instead of listing all five groupings, reason by where the addition happens. If the $+$ is done first or last, the two multiplications may be taken in either order without changing anything, giving one value each ($70$ and $26$). The only extra freedom is grouping the middle, which yields $\big((2\times3)+4\big)\times5=50$ and $2\times\big(3+(4\times5)\big)=46$. That is $4$ distinct values, confirming (C).
CCSS standards used (min grade 5)
5.OA.A.1Use parentheses, brackets, or braces in numerical expressions and evaluate (Reading how each set of parentheses changes the order of operations and evaluating each grouping from the inside out.)5.OA.A.2Write simple expressions that record calculations with numbers (Writing out the five full parenthesizations as separate numerical expressions to check.)3.OA.C.7Fluently multiply and divide within 100 (Computing the products such as $2\times7=14$, $7\times5=35$, and $14\times5=70$ inside each grouping.)
⭐ Parentheses only change which operation goes first, so list every grouping, work each one out, and count the answers that are actually different.
⭐ Parentheses only change which operation goes first, so list every grouping, work each one out, and count the answers that are actually different.
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