Sensim Math Original · sm-8
Easy mode Grade 4Problem
Imagine a greenhouse rack shaped like a 7×7 grid — 7 rows across and 7 columns down, so 49 little spots in all.
A florist named Yuna has 49 potted orchids. She labels them 1,2,3,…,49 (each number used once) and puts one orchid in each spot, filling the whole rack.
Now look at the rack one shelf at a time. Each row is a shelf with 7 orchids on it, and each column is also a shelf with 7 orchids on it. So in total there are 14 shelves to look at (7 rows plus 7 columns).
For each shelf, multiply together the 7 orchid numbers on it. Call that shelf lucky if its product is not a multiple of 7.
Yuna can choose any arrangement she likes. What is the greatest possible number of lucky shelves she can have?
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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