AMC 10 · 2023 · #15

Grade 8 arithmetic

perfect-squaresprime-factorizationfactorialparityexponents identify-subproblemspattern-recognitioncasework ↑ Prerequisites: prime-factorizationperfect-squaresfactorial

📏 Long solution 💡 3 insights

Problem

What is the least positive integer $m$ such that $m\cdot2!\cdot3!\cdot4!\cdot5!...16!$ is a perfect square?

$\textbf{(A) }30\qquad\textbf{(B) }30030\qquad\textbf{(C) }70\qquad\textbf{(D) }1430\qquad\textbf{(E) }1001$

Pick an answer.

(A)

(B)

30030

(C)

(D)

1430

(E)

1001

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Toolkit + CCSS Solution

Understand

Restated: Find the smallest positive integer $m$ that makes $m \cdot 2! \cdot 3! \cdot 4! \cdot 5! \cdots 16!$ a perfect square.

Givens: Product $N = 2! \cdot 3! \cdot 4! \cdot 5! \cdots 16!$ — fifteen factorials from $2!$ to $16!$; Want least positive integer $m$ such that $m \cdot N$ is a perfect square; Answer choices: (A) $30$, (B) $30030$, (C) $70$, (D) $1430$, (E) $1001$

Unknowns: Least positive integer $m$

Understand

Restated: Find the smallest positive integer $m$ that makes $m \cdot 2! \cdot 3! \cdot 4! \cdot 5! \cdots 16!$ a perfect square.

Plan

Primary tool: #7 Identify Subproblems

Secondary: #5 Look for a Pattern, #16 Change Focus / Count the Complement, #2 Make a Systematic List, #3 Eliminate Possibilities

Tool #7 (Identify Subproblems) splits the work: (a) reduce $N$ to a form that exposes a perfect-square chunk, (b) find the leftover non-square chunk, (c) determine which primes in the leftover have odd exponents — those are exactly what $m$ must contain. Tool #5 (Pattern) catches the pairing trick: $(2k)! \cdot (2k+1)! = (2k+1) \cdot [(2k)!]^2$, so consecutive pairs of factorials peel off a square. Tool #16 (Change Focus) shifts the question from "is $m N$ a perfect square?" (hard) to "which primes in $N$ have odd exponents?" (a clean parity check) — the complement of the perfect-square structure. Tool #2 (Systematic List) of primes $2, 3, 5, 7, 11, 13$ via Legendre's formula on $16!$ does the bookkeeping.

Execute — Answer: C

#5 Look for a Pattern 6.EE.A.3 Step 1

Pair adjacent factorials.
There are $15$ factorials from $2!$ to $16!$ — pair the first $14$ as $(2! \cdot 3!), (4! \cdot 5!), \ldots, (14! \cdot 15!)$ ($7$ pairs) and leave $16!$ alone.
For each pair, $(2k)! \cdot (2k+1)! = (2k)! \cdot (2k+1) \cdot (2k)! = (2k+1) \cdot [(2k)!]^2$, which is $(2k+1)$ times a perfect square.

$$(2k)! \cdot (2k+1)! = (2k+1) \cdot [(2k)!]^2$$

💡 Adjacent factorials share most of their factors — the pattern peels off a clean square — Grade 6 equivalent-expression move.

#7 Identify Subproblems 6.EE.A.3 Step 2

Apply the pattern to all $7$ pairs and isolate the squares.
The $7$ pairs contribute the leftover factors $3, 5, 7, 9, 11, 13, 15$ (the odd numbers from $3$ to $15$) and the perfect square block $[(2!)^2 \cdot (4!)^2 \cdots (14!)^2]$ which is already a square.
So $N = (3 \cdot 5 \cdot 7 \cdot 9 \cdot 11 \cdot 13 \cdot 15) \cdot \bigl(\text{perfect square}\bigr) \cdot 16!$.

$$N = (3 \cdot 5 \cdot 7 \cdot 9 \cdot 11 \cdot 13 \cdot 15) \cdot \text{square} \cdot 16!$$

💡 Once a perfect-square block is identified, ignore it — only the non-square chunk decides what $m$ must supply.

#2 Make a Systematic List 6.NS.B.4 Step 3

Find the parity of each prime's exponent in $K = (3 \cdot 5 \cdot 7 \cdot 9 \cdot 11 \cdot 13 \cdot 15)$.
Factor each: $9 = 3^2, 15 = 3 \cdot 5$.
So $K = 3^{1+2+1} \cdot 5^{1+1} \cdot 7 \cdot 11 \cdot 13 = 3^4 \cdot 5^2 \cdot 7 \cdot 11 \cdot 13$.
Even exponents: $3, 5$.
Odd exponents: $7, 11, 13$.

$$K = 3^4 \cdot 5^2 \cdot 7^1 \cdot 11^1 \cdot 13^1$$

💡 Factor each odd number into primes and tally — Grade 6 GCF / prime-factor reasoning.

#2 Make a Systematic List 8.EE.A.1 Step 4

Find the parity of each prime's exponent in $16!$ via Legendre's formula: $E_p(16!) = \lfloor 16/p \rfloor + \lfloor 16/p^2 \rfloor + \ldots$.
Primes $\le 16$ are $2, 3, 5, 7, 11, 13$.

$$E_2 = 8+4+2+1 = 15 \text{ (odd)}; \; E_3 = 5+1 = 6 \text{ (even)}; \; E_5 = 3 \text{ (odd)}; \; E_7 = 2 \text{ (even)}; \; E_{11} = 1 \text{ (odd)}; \; E_{13} = 1 \text{ (odd)}$$

💡 Legendre's formula counts multiples of $p$, $p^2$, $\ldots$ inside $16!$ — Grade 8 integer-exponent bookkeeping.

#16 Change Focus / Count the Complement 8.EE.A.1 Step 5

Combine $K$ and $16!$.
The exponent of each prime in $K \cdot 16!$ is the sum.
Only the parity matters for "perfect square".
Compute parities: $E_2: \text{(none)} + 15 = 15$ odd; $E_3: 4 + 6 = 10$ even; $E_5: 2 + 3 = 5$ odd; $E_7: 1 + 2 = 3$ odd; $E_{11}: 1 + 1 = 2$ even; $E_{13}: 1 + 1 = 2$ even.

$$K \cdot 16! = 2^{\text{odd}} \cdot 3^{\text{even}} \cdot 5^{\text{odd}} \cdot 7^{\text{odd}} \cdot 11^{\text{even}} \cdot 13^{\text{even}} \cdot \bigl(\text{square}\bigr)$$

💡 Switch focus from "the exponent" to "the exponent mod $2$" — Grade 8 parity check identifies exactly the primes $m$ must cover.

#7 Identify Subproblems 6.NS.B.4 Step 6

The primes with odd exponent in $N$ are $2, 5, 7$.
The smallest $m$ that makes every exponent even contributes one factor of each.
So $m = 2 \cdot 5 \cdot 7 = 70$.

$$m = 2 \cdot 5 \cdot 7 = 70$$

💡 Adding one of each odd-exponent prime flips them all to even — Grade 6 prime-factor matching.

#3 Eliminate Possibilities 6.NS.B.4 Step 7

Match to the choices: $70$ is (C).
The other choices come from omitting or mis-including primes — $30 = 2 \cdot 3 \cdot 5$ (wrongly drops $7$, adds $3$), $30030 = 2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 \cdot 13$ (extras), $1430 = 2 \cdot 5 \cdot 11 \cdot 13$ (wrong primes), $1001 = 7 \cdot 11 \cdot 13$ (drops $2, 5$).

$$m = 70 \;\Rightarrow\; \textbf{(C)}$$

💡 $70$ is the unique square-free $m$ covering exactly $\{2, 5, 7\}$ — multiple-choice closeout.

[1] #5 6.EE.A.3 Pair adjacent factorials. There are $15$ factorials from $2!$ to $16!$ — pair th

[2] #7 6.EE.A.3 Apply the pattern to all $7$ pairs and isolate the squares. The $7$ pairs contri

[3] #2 6.NS.B.4 Find the parity of each prime's exponent in $K = (3 \cdot 5 \cdot 7 \cdot 9 \cdo

[4] #2 8.EE.A.1 Find the parity of each prime's exponent in $16!$ via Legendre's formula: $E_p(1

[5] #16 8.EE.A.1 Combine $K$ and $16!$. The exponent of each prime in $K \cdot 16!$ is the sum. O

[6] #7 6.NS.B.4 The primes with odd exponent in $N$ are $2, 5, 7$. The smallest $m$ that makes e

[7] #3 6.NS.B.4 Match to the choices: $70$ is (C). The other choices come from omitting or mis-i

Review

Reasonableness: Cross-check the parity of $7$ in $N$. From $K$: $7$ appears once. From $16!$: $E_7(16!) = \lfloor 16/7 \rfloor + \lfloor 16/49 \rfloor = 2 + 0 = 2$. Total exponent of $7$ in $N$: $1 + 2 = 3$ — odd, so $m$ needs a factor of $7$. Consistent. Cross-check $11$: $K$ contributes $11^1$, $16!$ contributes $11^1$, total $11^2$ — even, $m$ does not need $11$. Consistent. Cross-check $2$: $K$ contributes $2^0$, $16!$ contributes $2^{15}$, total odd — $m$ needs $2$. Consistent. So $m = 2 \cdot 5 \cdot 7 = 70$ is square-free and minimal. Magnitude check: $70$ is much smaller than $30030 = 70 \cdot 11 \cdot 13 \cdot 3$, so $70$ being the smallest is reasonable.

Alternative: Tool #13 (Convert to Algebra) writes $N$ directly as a product of prime powers using Legendre on each $k!$ and sums up. The bookkeeping is heavier but produces the same odd-exponent set $\{2, 5, 7\}$. Slower because each factorial gets its own Legendre table; the pairing trick of Step 1 cuts the work in half.

CCSS standards used (min grade 8)

6.EE.A.3 Apply the properties of operations to generate equivalent expressions (Rewriting $(2k)! \cdot (2k+1)! = (2k+1) \cdot [(2k)!]^2$ to expose the perfect-square chunk.)
6.NS.B.4 Find greatest common factor and least common multiple of two numbers (Prime-factoring $K = 3 \cdot 5 \cdot 7 \cdot 9 \cdot 11 \cdot 13 \cdot 15$ into $3^4 \cdot 5^2 \cdot 7 \cdot 11 \cdot 13$, and building $m = 2 \cdot 5 \cdot 7$.)
8.EE.A.1 Know and apply the properties of integer exponents (Applying Legendre's formula to $16!$ and reasoning about the parity of each prime's exponent.)

⭐ This AMC 10 problem only needs Grade 8 "properties of integer exponents" you already know — pairing consecutive factorials peels off perfect squares, and the only primes left with odd exponents in the leftover are $2, 5, 7$, so the smallest $m$ is $2 \cdot 5 \cdot 7 = 70$.