Sensim Math Original · sm-8
SM Original Grade 4 number-theorycountingProblem
A florist named Yuna labels 49 potted orchids with the numbers 1,2,3,…,49, using each label exactly once, and stands the pots on a 7×7 greenhouse rack so that every row and every column holds 7 pots. For each of the 7 rows and each of the 7 columns, Yuna multiplies together the 7 orchid numbers on that shelf. A shelf is called lucky if its product is not a multiple of 7. What is the greatest possible number of lucky shelves Yuna can arrange?
Pick an answer.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: Yuna places the labels $1$ through $49$, one per pot, on a $7 \times 7$ rack of orchids. For every row and every column she multiplies the $7$ labels on that shelf. A shelf is **lucky** when its product is **not** divisible by $7$. We want the **largest** total number of lucky shelves (rows-plus-columns) over all possible arrangements.
Givens: A $7 \times 7$ rack with $49$ cells; one numbered orchid per cell; Labels $1, 2, 3, \ldots, 49$ are each used exactly once; There are $7$ rows and $7$ columns, so $14$ shelves total; Answer choices: (A) 6, (B) 7, (C) 8, (D) 9, (E) 10
Unknowns: The **maximum** total of (lucky rows) + (lucky columns), where a shelf is lucky iff its product is not a multiple of $7$
Understand
Restated: Yuna places the labels $1$ through $49$, one per pot, on a $7 \times 7$ rack of orchids. For every row and every column she multiplies the $7$ labels on that shelf. A shelf is **lucky** when its product is **not** divisible by $7$. We want the **largest** total number of lucky shelves (rows-plus-columns) over all possible arrangements.
Givens: A $7 \times 7$ rack with $49$ cells; one numbered orchid per cell; Labels $1, 2, 3, \ldots, 49$ are each used exactly once; There are $7$ rows and $7$ columns, so $14$ shelves total; Answer choices: (A) 6, (B) 7, (C) 8, (D) 9, (E) 10
Plan
Primary tool: #16 Change Focus / Complement
Secondary: #9 Solve an Easier Related Problem, #2 Make a Systematic List, #6 Guess and Check
Counting lucky shelves directly is awkward, so flip the question with Tool #16 (complement): each lucky shelf is one that does **not** contain any multiple of $7$, so maximizing lucky shelves is the same as **minimizing** the rows and columns that DO contain a multiple of $7$. That converts the puzzle into the much simpler question 'put the $7$ multiples of $7$ inside the smallest rows × columns rectangle'. Tool #9 then reduces it further to 'find positive integers $r, c \le 7$ with $r \times c \ge 7$ that make $r + c$ as small as possible'. Tool #2 lists those $(r, c)$ pairs and Tool #6 checks the smallest candidate sums. No algebra (#13) is needed.
Execute — Answer: C
4.OA.B.4 Step 1 - Begin with the key divisibility fact: a product of whole numbers is a multiple of $7$ if and only if at least one of those numbers is a multiple of $7$.
- So a shelf is lucky exactly when **none** of its $7$ labels is a multiple of $7$.
- By the complement (Tool #16), instead of maximizing lucky shelves we will minimize the shelves that DO contain a multiple of $7$.
- Specifically, if $r$ rows and $c$ columns contain at least one multiple of $7$, the number of lucky shelves is $(7 - r) + (7 - c) = 14 - (r + c)$, and we want to make $r + c$ as small as possible.
💡 Grade 4 factor / multiple thinking says a product picks up the factor $7$ only when one of the numbers already has it — flipping to 'count clean shelves' is just the complement.
3.OA.C.7 Step 2 - Count the multiples of $7$ among $1, 2, \ldots, 49$.
- They are $7, 14, 21, 28, 35, 42, 49$, and the largest one is $49 = 7 \times 7$, so the count is $49 \div 7 = 7$.
- Exactly $7$ of the $49$ cells on the rack hold a multiple of $7$.
💡 Dividing $49 \div 7 = 7$ is Grade 3 fluent multiplication and division within $100$.
3.MD.C.7 Step 3 - Reduce the problem to a smaller, cleaner one (Tool #9).
- Suppose those $7$ multiples of $7$ end up in $r$ different rows and $c$ different columns.
- Then all $7$ multiples sit inside the $r \times c$ rectangle of cells where those rows and columns meet, so that rectangle must have at least $7$ cells: $r \times c \ge 7$.
- The rack is $7 \times 7$, so $1 \le r, c \le 7$.
- The new question is: among positive integers $r, c \le 7$ with $r \times c \ge 7$, what is the smallest value of $r + c$?
💡 Grade 3 'area = rows $\times$ columns' says any $r$ chosen rows and $c$ chosen columns enclose an $r \times c$ block of cells — the natural cage for the seven multiples of $7$.
3.OA.C.7 Step 4 - Try the smallest sums first by guess-and-check (Tool #6).
- For sum $r + c = 5$, the largest product comes from splitting $5$ as evenly as possible: $(2, 3)$ gives $r \times c = 6 < 7$, and $(1, 4)$ gives $4 < 7$.
- So $r + c = 5$ cannot fit all $7$ multiples and is **impossible**.
💡 Quick Grade 3 mental multiplication ($2 \times 3 = 6$, compare with $7$) rules out the smallest sum.
4.OA.A.3 Step 5 - Next try $r + c = 6$, and systematically list every pair $(r, c)$ with $r, c \ge 1$ summing to $6$ (Tool #2): $(1, 5), (2, 4), (3, 3), (4, 2), (5, 1)$.
- Their products are $5, 8, 9, 8, 5$.
- The pairs whose product is at least $7$ are $(2, 4), (3, 3), (4, 2)$.
- So $r + c = 6$ is **achievable** — for example, place all $7$ multiples of $7$ inside some $3 \times 3$ sub-block of the rack (which has $9 \ge 7$ cells, so there is even room to spare), and fill the rest of the rack with the other $42$ labels.
💡 Listing every $(r, c)$ pair summing to $6$, multiplying each, and comparing with $7$ is a Grade 4 multi-step problem-solving move.
4.OA.A.3 Step 6 - Putting it together: $\min(r + c) = 6$, so the maximum number of lucky shelves is $14 - 6 = 8$.
- That matches choice (C).
- The other choices fail because the smaller sums (A) $6 = 14 - 8$ and (B) $7 = 14 - 7$ would need $r + c \ge 7$, which we beat; the larger choices (D) $9$ and (E) $10$ would need $r + c \le 5$, which Step 4 ruled out.
- So the unique answer is (C) $8$.
💡 Subtract the smallest possible contaminated count from $14$ — Grade 4 multi-step reasoning with the four operations.
4.OA.B.4 Begin with the key divisibility fact: a product of whole numbers is a multiple o 3.OA.C.7 Count the multiples of $7$ among $1, 2, \ldots, 49$. They are $7, 14, 21, 28, 35 3.MD.C.7 Reduce the problem to a smaller, cleaner one (Tool #9). Suppose those $7$ multip 3.OA.C.7 Try the smallest sums first by guess-and-check (Tool #6). For sum $r + c = 5$, t 4.OA.A.3 Next try $r + c = 6$, and systematically list every pair $(r, c)$ with $r, c \ge 4.OA.A.3 Putting it together: $\min(r + c) = 6$, so the maximum number of lucky shelves i Review
Reasonableness: Does $8$ make sense? The rack has $7 + 7 = 14$ shelves total, so the answer must lie between $0$ and $14$, and $8$ comfortably does. On the high side, $7$ multiples of $7$ must occupy at least $\lceil 7 / 7 \rceil = 1$ row and at least $1$ column, so at least $1 + 1 = 2$ shelves are contaminated and at most $14 - 2 = 12$ shelves are lucky. On the low side, packing all $7$ multiples into a $3 \times 3$ corner shows the bound $8$ is actually reached: $4$ lucky rows + $4$ lucky columns = $8$ lucky shelves. So $8$ sits between the rough upper bound $12$ and is itself achievable — it is the genuine maximum. A concrete construction: put $7, 14, 21, 28, 35, 42, 49$ into any $7$ cells of the $3 \times 3$ block formed by rows $1$–$3$ and columns $1$–$3$, then fill the remaining $42$ cells with the $42$ non-multiples of $7$ in any order.
Alternative: An alternative is Tool #2 (Make a Systematic List) without the complement flip: directly tabulate, for each candidate value of $r + c$ from $14$ down to $0$, whether the $7$ multiples of $7$ can possibly fit in the corresponding $r \times c$ rectangle. That route reaches the same inequality $r \times c \ge 7$ and the same minimum $r + c = 6$, hence the same answer $14 - 6 = 8$, but the complement-first framing (Tool #16) is faster because it tells you immediately what to minimize.
CCSS standards used (min grade 4)
3.OA.C.7Fluently multiply and divide within 100 (Computing $49 \div 7 = 7$ to count multiples of $7$, and checking small products like $2 \times 3 = 6$ and $3 \times 3 = 9$ against $7$.)3.MD.C.7Relate area to multiplication and addition operations (Seeing that any $r$ chosen rows and $c$ chosen columns of the rack enclose an $r \times c$ rectangle that must hold all $7$ multiples of $7$.)4.OA.B.4Find all factor pairs and recognize multiples; determine prime or composite (Using the fact that a product is a multiple of $7$ iff a factor is, and identifying the $7$ multiples of $7$ between $1$ and $49$.)4.OA.A.3Solve multi-step word problems using four operations with whole numbers (Listing every $(r, c)$ pair summing to small values, comparing each product with $7$, and combining via the complement to get $14 - \min(r + c)$.)
⭐ This AMC 8 problem only needs Grade 4 factors and multiples plus multi-step problem solving you already know!
⭐ This AMC 8 problem only needs Grade 4 factors and multiples plus multi-step problem solving you already know!
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