AMC 10 · 2022 · #7

Grade 6 number-theory

lcmgcdprime-factorizationdigit-sum systematic-enumerationcasework ↑ Prerequisites: prime-factorization

📏 Medium solution 💡 2 insights

Problem

The least common multiple of a positive integer $n$ and $18$ is $180$ , and the greatest common divisor of $n$ and $45$ is $15$ . What is the sum of the digits of $n$ ?

$\textbf{(A) } 3 \qquad \textbf{(B) } 6 \qquad \textbf{(C) } 8 \qquad \textbf{(D) } 9 \qquad \textbf{(E) } 12$

Pick an answer.

(A)

(B)

(C)

(D)

(E)

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Toolkit + CCSS Solution

Understand

Restated: A positive integer $n$ satisfies two conditions: the least common multiple of $n$ and $18$ is $180$, and the greatest common divisor of $n$ and $45$ is $15$. Find $n$ and add up its digits.

Givens: $\operatorname{lcm}(n, 18) = 180$; $\gcd(n, 45) = 15$; $n$ is a positive integer; Answer choices for the digit sum: (A) $3$, (B) $6$, (C) $8$, (D) $9$, (E) $12$

Unknowns: The value of $n$; The sum of the digits of $n$

Understand

Givens: $\operatorname{lcm}(n, 18) = 180$; $\gcd(n, 45) = 15$; $n$ is a positive integer; Answer choices for the digit sum: (A) $3$, (B) $6$, (C) $8$, (D) $9$, (E) $12$

Plan

Primary tool: #2 Make a Systematic List

Secondary: #3 Eliminate Possibilities

The two conditions each force a small, listable shape on $n$. Tool #2 (Systematic List): the GCD condition says $15 \mid n$, so write the multiples of $15$ up to $180$ in order. That short list is the entire candidate pool. Tool #3 (Eliminate Possibilities): now run each candidate through the two original conditions and keep the one survivor. This avoids the prime-exponent algebra of the reference solution while still being rigorous, and shows the student exactly why exactly one $n$ works.

Execute — Answer: B

#2 Make a Systematic List 4.OA.B.4 Step 1

The GCD condition $\gcd(n, 45) = 15$ implies $15$ divides $n$ (otherwise no common factor of $15$).
So $n$ is a multiple of $15$.
The LCM condition implies $n \mid 180$, so $n \le 180$.
List all multiples of $15$ up to $180$.

$$n \in \{15, 30, 45, 60, 75, 90, 105, 120, 135, 150, 165, 180\}$$

💡 Listing every multiple of $15$ in order guarantees no candidate is missed.

#3 Eliminate Possibilities 4.OA.B.4 Step 2

Apply the LCM condition first: $\operatorname{lcm}(n, 18) = 180$.
Since the LCM must be $180$, $n$ itself must divide $180$.
Cross out any candidate that does not divide $180$.
Divisors of $180$ among the list: $15, 30, 45, 60, 90, 180$.
Drop $75, 105, 120, 135, 150, 165$.

Survivors after $n \mid 180$: $\{15, 30, 45, 60, 90, 180\}$

💡 Any number whose LCM with anything equals $180$ must itself be a divisor of $180$.

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

Check each survivor's $\operatorname{lcm}(n, 18)$: $\operatorname{lcm}(15,18)=90$ ✗.
$\operatorname{lcm}(30,18)=90$ ✗.
$\operatorname{lcm}(45,18)=90$ ✗.
$\operatorname{lcm}(60,18)=180$ ✓.
$\operatorname{lcm}(90,18)=90$ ✗.
$\operatorname{lcm}(180,18)=180$ ✓.
Two pass: $n \in \{60, 180\}$.

$$\operatorname{lcm}(60,18)=180,\;\operatorname{lcm}(180,18)=180$$

💡 Computing $\operatorname{lcm}$ for a handful of small pairs is quick once the candidate pool is shrunk.

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

Now apply the GCD condition $\gcd(n, 45) = 15$ to the two survivors.
$\gcd(60, 45) = 15$ ✓ (since $60 = 4\cdot 15$ and $45 = 3\cdot 15$, share exactly the factor $15$).
$\gcd(180, 45) = 45$ ✗ (since $45 \mid 180$).
Only $n = 60$ passes both.

$$\gcd(60,45)=15\;\checkmark,\;\gcd(180,45)=45\;✗$$

💡 If $45$ already divides $n$, then $\gcd(n,45)=45$, not $15$ — so $n$ may be a multiple of $15$ but not of $45$.

#2 Make a Systematic List 2.NBT.A.1 Step 5

$n = 60$.
The digits of $60$ are $6$ and $0$.
Their sum is $6 + 0 = 6$, which is choice (B).

$$6 + 0 = 6\;\Rightarrow\;\textbf{(B)}$$

💡 Reading digits and adding them is the same skill as place value in second-grade arithmetic.

[1] #2 4.OA.B.4 The GCD condition $\gcd(n, 45) = 15$ implies $15$ divides $n$ (otherwise no comm

[2] #3 4.OA.B.4 Apply the LCM condition first: $\operatorname{lcm}(n, 18) = 180$. Since the LCM

[3] #3 6.NS.B.4 Check each survivor's $\operatorname{lcm}(n, 18)$: $\operatorname{lcm}(15,18)=90

[4] #3 6.NS.B.4 Now apply the GCD condition $\gcd(n, 45) = 15$ to the two survivors. $\gcd(60, 4

[5] #2 2.NBT.A.1 $n = 60$. The digits of $60$ are $6$ and $0$. Their sum is $6 + 0 = 6$, which is

Review

Reasonableness: Verify $n=60$ against both original conditions: $\operatorname{lcm}(60,18) = 180$ ✓ and $\gcd(60,45) = 15$ ✓. The digit sum $6+0=6$ matches choice (B), and the other choices $3, 8, 9, 12$ correspond to digit sums of $n=30, 35/53, 45/54/63, 39/48/57/75/84/93$ — none of which simultaneously satisfy both original constraints (e.g. $30$ fails because $\operatorname{lcm}(30,18)=90$ not $180$).

Alternative: Tool #13 (Convert to Algebra) via prime factorization. Write $n = 2^a 3^b 5^c$ (only primes that appear in $18, 45, 180, 15$). LCM condition $\operatorname{lcm}(n,18)=180=2^2 3^2 5$ forces $\max(a,1)=2$, $\max(b,2)=2$, $\max(c,0)=1$, so $a=2,\;b\le 2,\;c=1$. GCD condition $\gcd(n,45)=15=3\cdot 5$ forces $\min(b,2)=1$, so $b=1$. Thus $n=2^2\cdot 3\cdot 5 = 60$. Same answer, but the systematic-list path is friendlier for students who have not yet practiced the $\max/\min$ prime exponent rules.

CCSS standards used (min grade 6)

4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite (Listing multiples of $15$ up to $180$ and recognizing divisors of $180$.)
6.NS.B.4 Find greatest common factor and least common multiple of two numbers (Computing $\operatorname{lcm}(n,18)$ and $\gcd(n,45)$ for each candidate to filter to $n=60$.)
2.NBT.A.1 Understand that the three digits of a three-digit number represent hundreds, tens, and ones (Reading the digits $6$ and $0$ from $n=60$ and adding them.)

⭐ This AMC 10 problem only needs Grade 6 LCM and GCD reasoning you already know — list multiples of $15$, keep the ones that divide $180$, and check which one gives LCM $180$ and GCD $15$; that is $n=60$, with digit sum $6$.