AMC 10 · 2018 · #3
Easy mode Grade 3You have the four digits 1,2,3, and 4. Put one digit in each blank of (×)+(×), using each digit exactly once. Multiply the two digits in each pair, then add the two products. How many different totals can you get?
Pick an answer.
AMC 10 2018 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 four blanks in $(\,\underline{\;}\times\underline{\;}\,)+(\,\underline{\;}\times\underline{\;}\,)$ get the digits $1,2,3,4$, each used exactly once. Count how many different totals the expression can equal.
Givens: The form is (product) + (product): two pairs of digits, each pair multiplied, then the two products added; The digits $1,2,3,4$ are each used exactly once across the four blanks; Answer choices: (A) $2$, (B) $3$, (C) $4$, (D) $6$, (E) $24$
Unknowns: The number of different values the expression can take
Understand
Restated: The four blanks in $(\,\underline{\;}\times\underline{\;}\,)+(\,\underline{\;}\times\underline{\;}\,)$ get the digits $1,2,3,4$, each used exactly once. Count how many different totals the expression can equal.
Givens: The form is (product) + (product): two pairs of digits, each pair multiplied, then the two products added; The digits $1,2,3,4$ are each used exactly once across the four blanks; Answer choices: (A) $2$, (B) $3$, (C) $4$, (D) $6$, (E) $24$
Plan
Primary tool: #2 Make a Systematic List
Secondary: #15 Organize Information in More Ways, #3 Eliminate Possibilities
A "how many different values" question with a small finite setup points straight at Tool #2 (Make a Systematic List). But listing all $4! = 24$ ways to fill the blanks would be wasteful, so first apply Tool #15 (Organize Information in More Ways): because both $\times$ and $+$ are commutative, the only thing that matters is how the four digits split into two pairs. That reframing turns 24 messy arrangements into just a handful of pairings to list. Tool #3 (Eliminate Possibilities) then confirms the count against the answer choices.
Execute — Answer: B
3.OA.B.5 Step 1 Order inside and between products does not matter
- Inside one product, $a\times b = b\times a$, so the order of the two digits in a pair is irrelevant.
- Between the two products, $P+Q = Q+P$, so it does not matter which pair is written first.
- The total depends only on how the four digits are split into two pairs.
💡 Commutativity means rearranging factors or swapping the two products never changes the result, so all $24$ fillings collapse onto just the distinct pairings.
3.OA.A.3 Step 2 List the ways to split into two pairs
- Pin the digit $1$ and decide its partner: it can pair with $2$, $3$, or $4$.
- Whichever it picks, the remaining two digits form the other pair automatically.
- That gives exactly three splits, with no repeats and none missed.
💡 Fixing one digit and choosing its partner is the systematic rule that lists every pairing once and only once.
3.OA.C.7 Step 3 Multiply each pair
For each split, multiply the two digits in each pair to get the two products that will be added.
💡 Each pairing just needs two small single-digit multiplications before adding.
2.OA.B.2 Step 4 Add the two products in each split
Add the two products from each split to get that split's total value.
💡 Summing the two products turns each pairing into a single final number to compare.
K.MD.B.3 Step 5 Count the distinct totals
- The three totals are $14$, $11$, and $10$.
- They are all different, so there are three different values.
- This rules out the larger choices like $6$ or $24$ (which would assume order matters) and matches choice (B).
💡 Once each pairing gives one number, counting the different results is just sorting them into distinct buckets and counting the buckets.
3.OA.B.5 Inside one product, $a\times b = b\times a$, so the order of the two digits in a 3.OA.A.3 Pin the digit $1$ and decide its partner: it can pair with $2$, $3$, or $4$. Whi 3.OA.C.7 For each split, multiply the two digits in each pair to get the two products tha 2.OA.B.2 Add the two products from each split to get that split's total value. K.MD.B.3 The three totals are $14$, $11$, and $10$. They are all different, so there are Review
Reasonableness: The big choices $24$ and $6$ are traps: $24 = 4!$ counts every ordered filling, and $6$ over-counts pairings; both ignore that $\times$ and $+$ are commutative. There are genuinely only $3$ ways to split four digits into two pairs, and the three totals $14$, $11$, $10$ happen to be distinct, so no further merging occurs. Three distinct values, choice (B), is consistent.
Alternative: Tool #16 (Change Focus / Count the Complement): the products always use all of $1,2,3,4$, and $1\times2\times3\times4=24$ stays fixed, but the sum of two products is what varies. Note the split $\{1,4\},\{2,3\}$ pairs the extremes together and gives the smallest total $10$, while $\{1,2\},\{3,4\}$ keeps the two big digits together and gives the largest total $14$ — a quick way to see the totals must spread out into three different values.
CCSS standards used (min grade 3)
K.MD.B.3Classify objects into given categories and count the numbers in each (Sorting the three totals $14$, $11$, $10$ into distinct values and counting how many different ones there are.)2.OA.B.2Fluently add and subtract within 20 using mental strategies (Adding the two products in each split, e.g. $2+12=14$, $3+8=11$, $4+6=10$.)3.OA.A.3Solve multiplication and division word problems within 100 (Listing the three ways to split $1,2,3,4$ into two pairs by fixing $1$ and choosing its partner.)3.OA.B.5Apply properties of operations as strategies to multiply and divide (Using commutativity of $\times$ and $+$ to collapse all $24$ blank-fillings down to the distinct pairings.)3.OA.C.7Fluently multiply and divide within 100 (Computing each pair's product, e.g. $3\times 4=12$ and $2\times 4=8$.)
⭐ Because $\times$ and $+$ don't care about order, the only choice is how to split $1,2,3,4$ into two pairs — and there are just three ways, giving $14$, $11$, $10$.
⭐ Because $\times$ and $+$ don't care about order, the only choice is how to split $1,2,3,4$ into two pairs — and there are just three ways, giving $14$, $11$, $10$.
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