Sensim Math Original · sm-8

Easy mode Grade 4

Problem

Imagine a greenhouse rack shaped like a $7 \times 7$ 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 $1, 2, 3, \dots, 49$ (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?

$\textbf{(A) } 6\qquad\textbf{(B) } 7\qquad\textbf{(C) } 8\qquad\textbf{(D) } 9\qquad\textbf{(E) } 10$

Pick an answer.

(A)

(B)

(C)

(D)

(E)

View mode:

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

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

#16 Change Focus / Complement 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.

$$\text{lucky shelves} = (7 - r) + (7 - c) = 14 - (r + c)$$

💡 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.

#9 Solve an Easier Related Problem 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 $7$.

$$\text{multiples of } 7 \text{ from } 1 \text{ to } 49 = 49 \div 7 = 7$$

💡 Dividing $49 \div 7 = 7$ is Grade 3 fluent multiplication and division within $100$.

#9 Solve an Easier Related Problem 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 $r + c$?

$$r \times c \ge 7,\quad 1 \le r, c \le 7,\quad \text{minimize } r + c$$

💡 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 $7$.

#6 Guess and Check 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**.

$$r + c = 5 \;\Rightarrow\; \max(r \times c) = 2 \times 3 = 6 < 7$$

💡 Quick Grade 3 mental multiplication ($2 \times 3 = 6$, compare with $7$) rules out the smallest sum.

#2 Make a Systematic List 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.

$$r + c = 6:\ (2, 4) \to 8,\ (3, 3) \to 9,\ (4, 2) \to 8 \;\ge\; 7\ \checkmark$$

💡 Listing every $(r, c)$ pair summing to $6$, multiplying each, and comparing with $7$ is a Grade 4 multi-step problem-solving move.

#6 Guess and Check 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) $8$.

$$\text{max lucky shelves} = 14 - \min(r + c) = 14 - 6 = 8 \;\Rightarrow\; \textbf{(C)}$$

💡 Subtract the smallest possible contaminated count from $14$ — Grade 4 multi-step reasoning with the four operations.

[1] #16 4.OA.B.4 Begin with the key divisibility fact: a product of whole numbers is a multiple o

[2] #9 3.OA.C.7 Count the multiples of $7$ among $1, 2, \ldots, 49$. They are $7, 14, 21, 28, 35

[3] #9 3.MD.C.7 Reduce the problem to a smaller, cleaner one (Tool #9). Suppose those $7$ multip

[4] #6 3.OA.C.7 Try the smallest sums first by guess-and-check (Tool #6). For sum $r + c = 5$, t

[5] #2 4.OA.A.3 Next try $r + c = 6$, and systematically list every pair $(r, c)$ with $r, c \ge

[6] #6 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.7 Fluently 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.7 Relate 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.4 Find 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.3 Solve 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!

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: C

Review

CCSS standards used (min grade 4)

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