AMC 10 · 2005 · #21
Grade 7 arithmeticFor how many positive integers n does 1+2+⋯+n evenly divide 6n?
Pick an answer.
AMC 10 2005 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: Add up the whole numbers from $1$ to $n$. For how many positive whole numbers $n$ does that running total divide $6n$ with no remainder?
Givens: The running total is $1+2+3+\dotsb+n$; We compare it against $6n$; "Evenly divide" means $6n$ divided by the total leaves no remainder; $n$ ranges over the positive integers $1, 2, 3, \dotsc$; Answer choices: (A) $3$, (B) $5$, (C) $7$, (D) $9$, (E) $11$
Unknowns: The count of positive integers $n$ for which the total divides $6n$ exactly
Understand
Restated: Add up the whole numbers from $1$ to $n$. For how many positive whole numbers $n$ does that running total divide $6n$ with no remainder?
Givens: The running total is $1+2+3+\dotsb+n$; We compare it against $6n$; "Evenly divide" means $6n$ divided by the total leaves no remainder; $n$ ranges over the positive integers $1, 2, 3, \dotsc$; Answer choices: (A) $3$, (B) $5$, (C) $7$, (D) $9$, (E) $11$
Plan
Primary tool: #4 Introduce a Variable
Secondary: #2 Make a Systematic List, #3 Eliminate Possibilities
The word "total from $1$ to $n$" is awkward to test one value at a time, so Tool #4 (Introduce a Variable) leads: replace the sum with the closed formula $\tfrac{n(n+1)}{2}$ so the whole question becomes a single expression in $n$. Dividing $6n$ by that expression and simplifying collapses the messy ratio down to $\tfrac{12}{n+1}$ — now the problem is just "when is $\tfrac{12}{n+1}$ a whole number?" That turns it into a factor hunt, so Tool #2 (Make a Systematic List) is used to list every factor of $12$, and Tool #3 (Eliminate Possibilities) throws out the one value that breaks the rule $n \ge 1$. What is left is the count.
Execute — Answer: B
6.EE.B.6 Step 1 Replace the sum with a formula
- Adding $1+2+3+\dotsb+n$ every time would be slow, so use the known shortcut for that running total: $1+2+\dotsb+n = \tfrac{n(n+1)}{2}$.
- (Pair the first and last terms, the second and second-to-last, and so on: each pair adds to $n+1$, and there are $\tfrac{n}{2}$ pairs.) Now the total is a single expression in $n$.
💡 One clean formula stands in for the whole growing sum, so I can work with $n$ directly instead of adding term by term.
6.NS.A.1 Step 2 Divide 6n by the total
- "Evenly divides" means the quotient $6n \div \tfrac{n(n+1)}{2}$ is a whole number.
- Dividing by a fraction is the same as multiplying by its flip, so $6n \div \tfrac{n(n+1)}{2} = 6n \cdot \tfrac{2}{n(n+1)} = \tfrac{12n}{n(n+1)}$.
💡 Flip-and-multiply turns the awkward "divide by a fraction" into an ordinary product I can simplify.
7.EE.A.1 Step 3 Cancel n to reveal the real condition
- Both the top and the bottom of $\tfrac{12n}{n(n+1)}$ carry a factor of $n$, and $n \ge 1$ so $n \ne 0$; cancel it to get $\tfrac{12}{n+1}$.
- The original total divides $6n$ evenly exactly when this leftover fraction $\tfrac{12}{n+1}$ is a whole number, and $\tfrac{12}{n+1}$ is a whole number exactly when $n+1$ is a factor of $12$.
💡 Cancelling the shared $n$ strips the problem down to a plain divisibility question about $n+1$.
4.OA.B.4 Step 4 List every factor of 12
- Write out all the positive factors of $12$ in order: $1, 2, 3, 4, 6, 12$ — six of them.
- Each one is a possible value of $n+1$, so subtract $1$ from each to get the matching $n$: from $n+1 = 1,2,3,4,6,12$ we get $n = 0, 1, 2, 3, 5, 11$.
💡 Every factor of $12$ hands me one candidate for $n$, so listing the factors lists the candidates.
6.EE.B.5 Step 5 Drop n = 0 and count
- The problem asks for positive integers, so $n = 0$ is not allowed and is thrown out; the factor $n+1 = 1$ cannot be used.
- Every other value works.
- That leaves $n = 1, 2, 3, 5, 11$ — five values.
- So the answer is (B).
💡 Only the value that breaks the "$n$ is positive" rule gets removed, and counting what survives gives the answer.
6.EE.B.6 Adding $1+2+3+\dotsb+n$ every time would be slow, so use the known shortcut for 6.NS.A.1 "Evenly divides" means the quotient $6n \div \tfrac{n(n+1)}{2}$ is a whole numbe 7.EE.A.1 Both the top and the bottom of $\tfrac{12n}{n(n+1)}$ carry a factor of $n$, and 4.OA.B.4 Write out all the positive factors of $12$ in order: $1, 2, 3, 4, 6, 12$ — six o 6.EE.B.5 The problem asks for positive integers, so $n = 0$ is not allowed and is thrown Review
Reasonableness: Spot-check the endpoints of the list by hand. For $n = 1$: the total is $1$ and $6n = 6$, and $1$ divides $6$. For $n = 11$: the total is $\tfrac{11\cdot 12}{2} = 66$ and $6n = 66$, and $66$ divides $66$. For $n = 5$: the total is $15$ and $6n = 30$, and $15$ divides $30$. Now check that a skipped value fails, say $n = 4$: the total is $10$ and $6n = 24$, and $24 \div 10 = 2.4$ is not whole — correctly excluded because $n+1 = 5$ is not a factor of $12$. The five survivors all check out, so $5$ is right.
Alternative: Skip the formula and just test small $n$ directly, watching for when the total outgrows $6n$. Compute the total $T$ and $6n$ for $n = 1, 2, 3, \dotsc$: $n=1$ gives $T=1 \mid 6$ (yes), $n=2$ gives $T=3 \mid 12$ (yes), $n=3$ gives $T=6 \mid 18$ (yes), $n=4$ gives $T=10 \nmid 24$ (no), $n=5$ gives $T=15 \mid 30$ (yes), and once $n \ge 6$ the total $\tfrac{n(n+1)}{2}$ climbs past $6n$ except when $n+1$ is a large factor of $12$, which only happens again at $n=11$ ($T=66 \mid 66$). Collecting the yes-cases gives the same five values, choice (B).
CCSS standards used (min grade 7)
6.EE.B.6Use variables to represent numbers and write expressions to solve problems (Replacing the running sum $1+2+\dotsb+n$ with the single expression $\tfrac{n(n+1)}{2}$ in the variable $n$.)6.NS.A.1Interpret and compute quotients of fractions and solve word problems (Dividing $6n$ by the fraction $\tfrac{n(n+1)}{2}$ using flip-and-multiply to get $\tfrac{12n}{n(n+1)}$.)7.EE.A.1Apply properties of operations to add, subtract, factor, and expand linear expressions (Cancelling the common factor $n$ to reduce $\tfrac{12n}{n(n+1)}$ to $\tfrac{12}{n+1}$.)4.OA.B.4Find all factor pairs and recognize multiples; determine prime or composite (Listing all positive factors of $12$ to find every allowed value of $n+1$.)6.EE.B.5Understand solving an equation or inequality as a process of finding values (Keeping only the values of $n$ that satisfy $n \ge 1$ and counting them.)
⭐ Swap the sum for $\tfrac{n(n+1)}{2}$, and the whole problem shrinks to "$n+1$ must be a factor of $12$" — count those and you are done.
⭐ Swap the sum for $\tfrac{n(n+1)}{2}$, and the whole problem shrinks to "$n+1$ must be a factor of $12$" — count those and you are done.
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