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AMC 8 · 2015 · #22

Grade 6 number-theory
Archetype Small-Domain Constrained Search
divisor-countprime-factorizationlcmfactorsmultiples systematic-enumerationcasework ↑ Prerequisites: factorsmultiplesdivisibility-rules
📏 Medium solution 💡 3 insights

Problem

On June 1, a group of students is standing in rows, with 15 students in each row. On June 2, the same group is standing with all of the students in one long row. On June 3, the same group is standing with just one student in each row. On June 4, the same group is standing with 6 students in each row. This process continues through June 12 with a different number of students per row each day. However, on June 13, they cannot find a new way of organizing the students. What is the smallest possible number of students in the group?

(A) 21(B) 30(C) 60(D) 90(E) 1080\textbf{(A) } 21 \qquad \textbf{(B) } 30 \qquad \textbf{(C) } 60 \qquad \textbf{(D) } 90 \qquad \textbf{(E) } 1080

Pick an answer.

(A)
21
(B)
30
(C)
60
(D)
90
(E)
1080
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Toolkit + CCSS Solution

Understand

Restated: A group of students stands in equal rows. On June 1 there are $15$ per row, on June 2 they form one long row (everyone in a single row), on June 3 they stand one per row, and on June 4 there are $6$ per row. They keep finding a *new* number of students per row every day through June 12, but on June 13 they run out of new arrangements. What is the smallest possible number of students?

Givens: An arrangement with $k$ students per row works exactly when $k$ divides $N$ (the group size); Specific arrangements that must work: $15$, $N$ itself, $1$, and $6$ students per row; From June 1 to June 12 they find $12$ different values of $k$ — so $N$ has at least $12$ divisors; On June 13 no new value works — so $N$ has at most $12$ divisors; Answer choices: (A) $21$, (B) $30$, (C) $60$, (D) $90$, (E) $1080$

Unknowns: The smallest possible value of $N$ (the number of students)

Understand

Restated: A group of students stands in equal rows. On June 1 there are $15$ per row, on June 2 they form one long row (everyone in a single row), on June 3 they stand one per row, and on June 4 there are $6$ per row. They keep finding a *new* number of students per row every day through June 12, but on June 13 they run out of new arrangements. What is the smallest possible number of students?

Givens: An arrangement with $k$ students per row works exactly when $k$ divides $N$ (the group size); Specific arrangements that must work: $15$, $N$ itself, $1$, and $6$ students per row; From June 1 to June 12 they find $12$ different values of $k$ — so $N$ has at least $12$ divisors; On June 13 no new value works — so $N$ has at most $12$ divisors; Answer choices: (A) $21$, (B) $30$, (C) $60$, (D) $90$, (E) $1080$

Plan

Primary tool: #3 Eliminate Possibilities

Secondary: #2 Make a Systematic List

Each day's arrangement corresponds to one divisor of $N$, so "$12$ days but no $13$th" simply says $N$ has exactly $12$ divisors. That gives us a single sharp test. We then use Tool #3 (Eliminate Possibilities) on the five answer choices: knock out any that fail the divisibility constraints ($15 \mid N$ and $6 \mid N$, i.e. $30 \mid N$) and any whose divisor count isn't $12$. To check the divisor count we use Tool #2 (Make a Systematic List) — list every divisor of each surviving candidate in order from smallest to largest and count them. Working from the smallest choice upward, the first one that passes both tests is the answer.

Execute — Answer: C

#3 Eliminate Possibilities 4.OA.B.4 Step 1
  • Translate the story into divisor language.
  • "$k$ students per row in $r$ rows" means $N = r \times k$, so $k$ must be a *divisor* of $N$.
  • The number of valid daily arrangements equals the number of positive divisors of $N$.
$$N = r \times k \;\Longleftrightarrow\; k \mid N$$

💡 Rectangular arrays match factor pairs — exactly the Grade 4 "factors and multiples" idea.

#3 Eliminate Possibilities 4.OA.B.4 Step 2
  • Read off the divisor count.
  • They find new arrangements for $12$ days (June 1-12) and then run out on June 13.
  • So $N$ has exactly $12$ positive divisors — no more, no less.
$$d(N) = 12$$

💡 "$12$ work, $13$th doesn't" is the most direct way the problem tells us the divisor count.

#3 Eliminate Possibilities 6.NS.B.4 Step 3
  • Eliminate choices that fail the divisibility constraints.
  • Since $15$ and $6$ are both divisors, $\operatorname{lcm}(15, 6) = 30$ must divide $N$.
  • Choice (A) $21$ is not a multiple of $30$ (it's not even a multiple of $15$), so cross it out.
  • The survivors are $30, 60, 90, 1080$.
$\operatorname{lcm}(15, 6) = 30,\; \text{so } 30 \mid N$. Cross out (A) $21$.

💡 $\operatorname{lcm}$ is the Grade 6 way to combine two divisibility requirements into one.

#2 Make a Systematic List 4.OA.B.4 Step 4
  • Test (B) $30$ by listing every divisor in order.
  • Pair each small divisor with its partner: $1 \cdot 30, 2 \cdot 15, 3 \cdot 10, 5 \cdot 6$.
Divisors of $30$: $1, 2, 3, 5, 6, 10, 15, 30$ — count $= 8$, not $12$. Cross out (B).

💡 Pairing $k$ with $N / k$ is the systematic way to list divisors without missing any.

#2 Make a Systematic List 4.OA.B.4 Step 5

Test (C) $60$ by listing divisors the same way: $1 \cdot 60, 2 \cdot 30, 3 \cdot 20, 4 \cdot 15, 5 \cdot 12, 6 \cdot 10$.

Divisors of $60$: $1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60$ — count $= 12$. ✓

💡 Six factor pairs $\Rightarrow$ exactly $12$ divisors — and $15$ and $6$ are both on the list.

#3 Eliminate Possibilities 4.OA.B.4 Step 6

$60$ passes both tests and is smaller than $90$ and $1080$, so it is the smallest valid choice.

Answer: $\boxed{60} \;\Rightarrow\; \textbf{(C)}$

💡 Once a candidate passes from the smallest end of the list, larger candidates are no longer minimal.

[1] #3 4.OA.B.4 Translate the story into divisor language. "$k$ students per row in $r$ rows" me
[2] #3 4.OA.B.4 Read off the divisor count. They find new arrangements for $12$ days (June 1-12)
[3] #3 6.NS.B.4 Eliminate choices that fail the divisibility constraints. Since $15$ and $6$ are
[4] #2 4.OA.B.4 Test (B) $30$ by listing every divisor in order. Pair each small divisor with it
[5] #2 4.OA.B.4 Test (C) $60$ by listing divisors the same way: $1 \cdot 60, 2 \cdot 30, 3 \cdot
[6] #3 4.OA.B.4 $60$ passes both tests and is smaller than $90$ and $1080$, so it is the smalles

Review

Reasonableness: Double-check that $60$ matches every clue: $60 / 15 = 4$ rows (June 1 works), $60 / 1 = 60$ rows (June 2 works), $60 / 60 = 1$ row (June 3 works), $60 / 6 = 10$ rows (June 4 works). The full divisor list $\{1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60\}$ has exactly $12$ entries, so June 1-12 each get a new arrangement and June 13 has nothing left. Every condition holds.

Alternative: Tool #8 (Analyze the Units — here, units of *exponents* in prime factorization) gives a fast formula. If $N = p_1^{a_1} p_2^{a_2} \cdots$, then $d(N) = (a_1 + 1)(a_2 + 1) \cdots$. We want $d(N) = 12 = 2 \cdot 2 \cdot 3$ with $30 = 2 \cdot 3 \cdot 5$ dividing $N$. The cheapest fit is $N = 2^2 \cdot 3^1 \cdot 5^1 = 60$, giving $(2+1)(1+1)(1+1) = 12$. Same answer (C).

CCSS standards used (min grade 6)

  • 4.OA.B.4 Find all factor pairs for a whole number; recognize multiples and factors (Connecting the rectangular row arrangements to divisors of $N$, and listing all $8$ divisors of $30$ and all $12$ divisors of $60$ via factor pairs.)
  • 6.NS.B.4 Find the greatest common factor and least common multiple of two whole numbers (Combining the two divisibility requirements $15 \mid N$ and $6 \mid N$ into the single requirement $\operatorname{lcm}(15, 6) = 30 \mid N$ to eliminate choice (A).)

⭐ This AMC 8 problem only needs Grade 6 factor-and-multiple reasoning: turn "rows" into divisors, then test the choices!

⭐ This AMC 8 problem only needs Grade 6 factor-and-multiple reasoning: turn "rows" into divisors, then test the choices!

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