AMC 10 · 2003 · #18
Grade 6 number-theoryWhat is the largest integer that is a divisor of
(n+1)(n+3)(n+5)(n+7)(n+9)
for all positive even integers n?
Pick an answer.
AMC 10 2003 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: The five factors $n+1,\,n+3,\,n+5,\,n+7,\,n+9$ multiply to one number. As $n$ runs through the positive even integers $2,4,6,\dots$, this product keeps changing. Find the largest whole number that divides the product no matter which positive even $n$ you pick.
Givens: The product is $(n+1)(n+3)(n+5)(n+7)(n+9)$.; $n$ is any positive even integer ($2, 4, 6, \dots$).; The divisor must work for every such $n$, not just some.; Answer choices: (A) $3$, (B) $5$, (C) $11$, (D) $15$, (E) $165$.
Unknowns: The largest integer that divides $(n+1)(n+3)(n+5)(n+7)(n+9)$ for all positive even $n$.
Understand
Restated: The five factors $n+1,\,n+3,\,n+5,\,n+7,\,n+9$ multiply to one number. As $n$ runs through the positive even integers $2,4,6,\dots$, this product keeps changing. Find the largest whole number that divides the product no matter which positive even $n$ you pick.
Givens: The product is $(n+1)(n+3)(n+5)(n+7)(n+9)$.; $n$ is any positive even integer ($2, 4, 6, \dots$).; The divisor must work for every such $n$, not just some.; Answer choices: (A) $3$, (B) $5$, (C) $11$, (D) $15$, (E) $165$.
Plan
Primary tool: #3 Eliminate Possibilities
Secondary: #5 Look for a Pattern, #6 Guess and Check, #15 Organize Information in More Ways
The choices are a fixed short list, so the goal is to keep the largest one that always divides and rule out the rest (Tool #3). First reorganize the expression: since $n$ is even, the five factors are just five consecutive odd integers (Tool #15), which turns the whole question into a fact about odd numbers in a row. Then look for the repeating structure of multiples (Tool #5): among five consecutive odds there is always exactly one multiple of $5$ and always at least one multiple of $3$, so $3$ and $5$ — and hence $15$ — divide every product. The only bigger choices ($11$ and $165$) both need a factor of $11$, so test a single well-chosen even value (Tool #6): if one product has no multiple of $11$, both are eliminated. What survives and is largest is the answer.
Execute — Answer: D
2.OA.C.3 Step 1 See five consecutive odd numbers
- Since $n$ is even, adding an odd number keeps it odd, so $n+1, n+3, n+5, n+7, n+9$ are all odd.
- They go up by $2$ each time, so they are five consecutive odd integers.
- For example $n=2$ gives $3,5,7,9,11$ and $n=8$ gives $9,11,13,15,17$.
- So the real question is: what always divides the product of five consecutive odd numbers?
💡 Even plus odd is odd, so the five factors are just five odd numbers in a row.
4.OA.B.4 Step 2 One factor is always a multiple of 5
- Look at the remainders of five consecutive odd numbers when divided by $5$.
- Each step adds $2$, so the remainders run through $0, 2, 4, 1, 3$ in some order — all five different remainders.
- One of them is always $0$, so exactly one of the five numbers is a multiple of $5$.
- Check: $3,5,7,9,11$ has $5$; $7,9,11,13,15$ has $15$.
- So $5$ divides the product for every even $n$.
💡 Five odd numbers spaced by $2$ hit all five possible remainders mod $5$, so one must land on a multiple of $5$.
4.OA.B.4 Step 3 One factor is always a multiple of 3
- Do the same with $3$.
- Among three consecutive odd numbers the remainders mod $3$ are all different (each step adds $2\equiv-1$, giving $0,2,1$ in some order), so at least one is a multiple of $3$ — and five in a row certainly still contain one.
- Check: $5,7,9,11,13$ has $9$; $11,13,15,17,19$ has $15$.
- So $3$ also divides every product.
- Since $3$ and $5$ have no common factor, their product $15$ divides the product for every even $n$.
💡 Three odd numbers in a row already cover all remainders mod $3$, so a multiple of $3$ is always in the list.
3.OA.C.7 Step 4 Test 11 with one example
- Now the only bigger choices, $11$ and $165 = 3\cdot5\cdot11$, both need a factor of $11$.
- A guaranteed divisor must work for every even $n$, so one bad example is enough to rule it out.
- Take $n=12$: the factors are $13, 15, 17, 19, 21$.
- Dividing each by $11$ leaves a remainder — none is a multiple of $11$ (the nearest multiples are $11$ and $22$).
- So $11$ does not divide the product for every even $n$, and neither does $165$.
💡 One example with no multiple of $11$ proves $11$ can't be a guaranteed divisor.
6.NS.B.4 Step 5 Keep the largest divisor that always works
- Compare the choices.
- Both $3$ (A) and $5$ (B) always divide, but they are smaller than $15$.
- Both $11$ (C) and $165$ (E) need a factor of $11$, which the $n=12$ example killed.
- That leaves $15$ (D): it always divides (from $3$ and $5$ together) and it is the largest surviving choice.
- So the largest integer dividing the product for all positive even $n$ is $15$, which is (D).
💡 The answer is the biggest number every product shares, and that common part is $3\times5=15$.
2.OA.C.3 Since $n$ is even, adding an odd number keeps it odd, so $n+1, n+3, n+5, n+7, n+ 4.OA.B.4 Look at the remainders of five consecutive odd numbers when divided by $5$. Each 4.OA.B.4 Do the same with $3$. Among three consecutive odd numbers the remainders mod $3$ 3.OA.C.7 Now the only bigger choices, $11$ and $165 = 3\cdot5\cdot11$, both need a factor 6.NS.B.4 Compare the choices. Both $3$ (A) and $5$ (B) always divide, but they are smalle Review
Reasonableness: $15 = 3\cdot5$, and we showed a multiple of $3$ and a multiple of $5$ always sit among the five odd factors, so $15$ genuinely divides every product — not just a lucky fit. It is also the largest choice that can: $11$ and $165$ die at $n=12$ (factors $13,15,17,19,21$, no multiple of $11$), while $3$ and $5$ are smaller than $15$. Nothing above $15$ is forced either — $n=10$ gives $11\cdot13\cdot15\cdot17\cdot19$, whose only factor of $3$ and only factor of $5$ both come from the single $15$, so no second $3$ or $5$ is guaranteed. That pins the answer at exactly $15$.
Alternative: Eliminate purely with two test cases. Any answer must divide the first case $n=2$, namely $3\cdot5\cdot7\cdot9\cdot11$; every choice $3,5,11,15,165$ divides this, so it can't separate them alone. But the answer must also divide $n=12$, that is $13\cdot15\cdot17\cdot19\cdot21$; here $11$ (and hence $165$) fails, while $3,5,15$ all still divide. The largest survivor is $15$, matching the direct proof. This leans on Tool #6 (test specific cases) feeding Tool #3 (eliminate).
CCSS standards used (min grade 6)
2.OA.C.3Determine whether a group of objects has an odd or even number (Recognizing that even $n$ makes all five factors odd, so they are five consecutive odd integers.)4.OA.B.4Find all factor pairs and recognize multiples; determine prime or composite (Showing five consecutive odd numbers always contain a multiple of $5$ and a multiple of $3$, so $15$ always divides.)3.OA.C.7Fluently multiply and divide within 100 (Dividing $13,15,17,19,21$ by $11$ to confirm none is a multiple of $11$.)6.NS.B.4Find greatest common factor and least common multiple of two numbers (Reading the problem as the greatest common divisor shared by every product and identifying it as $3\times5=15$.)
⭐ Five odd numbers in a row always hide a multiple of $3$ and a multiple of $5$, so $15$ always divides — but $11$ can go missing, so nothing bigger is guaranteed.
⭐ Five odd numbers in a row always hide a multiple of $3$ and a multiple of $5$, so $15$ always divides — but $11$ can go missing, so nothing bigger is guaranteed.
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