AMC 10 · 2022 · #7
Grade 6 number-theoryProblem
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?
Pick an answer.
AMC 10 2022 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
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
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) $12$
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
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 $180$.
💡 Listing every multiple of $15$ in order guarantees no candidate is missed.
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 $75, 105, 120, 135, 150, 165$.
💡 Any number whose LCM with anything equals $180$ must itself be a divisor of $180$.
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: $n \in \{60, 180\}$.
💡 Computing $\operatorname{lcm}$ for a handful of small pairs is quick once the candidate pool is shrunk.
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.
💡 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.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).
💡 Reading digits and adding them is the same skill as place value in second-grade arithmetic.
4.OA.B.4 The GCD condition $\gcd(n, 45) = 15$ implies $15$ divides $n$ (otherwise no comm 4.OA.B.4 Apply the LCM condition first: $\operatorname{lcm}(n, 18) = 180$. Since the LCM 6.NS.B.4 Check each survivor's $\operatorname{lcm}(n, 18)$: $\operatorname{lcm}(15,18)=90 6.NS.B.4 Now apply the GCD condition $\gcd(n, 45) = 15$ to the two survivors. $\gcd(60, 4 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.4Find 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.4Find 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.1Understand 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 $6$.
⭐ 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 $6$.