AMC 10 · 2003 · #25
Grade 6 number-theoryLet n be a 5-digit number, and let q and r be the quotient and the remainder, respectively, when n is divided by 100. For how many values of n is q+r divisible by 11?
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: A number $n$ has $5$ digits. Divide $n$ by $100$: call the quotient $q$ and the remainder $r$. Count how many of the $5$-digit numbers $n$ make the sum $q+r$ a multiple of $11$.
Givens: $n$ is a $5$-digit number, so $10000 \le n \le 99999$.; $q$ is the quotient and $r$ is the remainder when $n$ is divided by $100$, so $n = 100q + r$ with $0 \le r \le 99$.; We want the count of $n$ for which $q + r$ is divisible by $11$.; Answer choices: (A) $8180$, (B) $8181$, (C) $8182$, (D) $9000$, (E) $9090$.
Unknowns: The number of $5$-digit values of $n$ for which $11$ divides $q+r$.
Understand
Restated: A number $n$ has $5$ digits. Divide $n$ by $100$: call the quotient $q$ and the remainder $r$. Count how many of the $5$-digit numbers $n$ make the sum $q+r$ a multiple of $11$.
Givens: $n$ is a $5$-digit number, so $10000 \le n \le 99999$.; $q$ is the quotient and $r$ is the remainder when $n$ is divided by $100$, so $n = 100q + r$ with $0 \le r \le 99$.; We want the count of $n$ for which $q + r$ is divisible by $11$.; Answer choices: (A) $8180$, (B) $8181$, (C) $8182$, (D) $9000$, (E) $9090$.
Plan
Primary tool: #4 Introduce a Variable
Secondary: #9 Solve an Easier Related Problem, #5 Look for a Pattern
The clean way in is to name the pieces: Tool #4 (Introduce a Variable) writes the split as $n = 100q + r$, turning the word 'quotient and remainder' into one equation. The magic is what that equation reveals: rewrite $100q + r$ as $99q + (q + r)$. Since $99$ is a multiple of $11$, the messy target $q + r$ and the number $n$ itself always leave the same remainder when divided by $11$ (Tool #5, Look for a Pattern — the pattern is that $100$ is just $1$ more than a multiple of $11$). So asking '$11$ divides $q+r$' is exactly the same as asking '$11$ divides $n$.' That swap is Tool #9 (Solve an Easier Related Problem): instead of chasing two variables, we just count the $5$-digit multiples of $11$, which is a short, direct count.
Execute — Answer: B
4.NBT.B.6 Step 1 Write the division as one equation
- Dividing $n$ by $100$ gives a quotient $q$ and a remainder $r$, which means $n = 100q + r$ where the remainder satisfies $0 \le r \le 99$.
- Concretely, $q$ is the number formed by the first three digits of $n$ and $r$ is the number formed by its last two digits.
- For a $5$-digit $n$ from $10000$ to $99999$, the quotient $q$ runs from $100$ to $999$ and the remainder $r$ runs from $0$ to $99$.
💡 Dividing by $100$ just slices a number's last two digits off as the remainder and keeps the front as the quotient.
6.EE.A.3 Step 2 Rewrite so $q+r$ appears
- We care about $q + r$, but the equation shows $100q + r$.
- Split the $100q$ into $99q + q$ so that a copy of $q$ joins the $r$: $n = 100q + r = 99q + q + r = 99q + (q + r)$.
- Now the exact quantity we want, $q + r$, sits right inside the expression for $n$.
💡 Peeling $99q$ off of $100q$ leaves exactly one extra $q$ to pair with $r$, surfacing the sum we care about.
4.OA.B.4 Step 3 See that $q+r$ and $n$ share divisibility by $11$
- The term $99q$ is always a multiple of $11$, because $99 = 9 \times 11$.
- So $n = 99q + (q + r)$ is a multiple of $11$ plus the amount $(q + r)$.
- Adding a multiple of $11$ never changes whether a number is divisible by $11$.
- Therefore $11$ divides $q + r$ exactly when $11$ divides $n$ itself.
- The hard-looking condition on $q + r$ is really just the condition that $n$ is a multiple of $11$.
💡 Two numbers that differ by a multiple of $11$ always give the same answer to 'is it divisible by $11$?'
4.OA.B.4 Step 4 Turn it into counting multiples of 11
- The question is now simply: how many $5$-digit numbers are multiples of $11$?
- Find the smallest and largest.
- The smallest $5$-digit multiple of $11$: $10000 \div 11 \approx 909.1$, so the next whole multiple is $11 \times 910 = 10010$.
- The largest: $99999 \div 11 \approx 9090.8$, so the last whole multiple is $11 \times 9090 = 99990$.
- Every multiple of $11$ from $10010$ up to $99990$ is a valid $n$.
💡 The valid values of $n$ are an evenly spaced list of multiples of $11$, pinned between the first and last one that still has five digits.
4.OA.C.5 Step 5 Count the list
- The multiples of $11$ from $10010$ to $99990$ are $11 \times 910, 11 \times 911, \ldots, 11 \times 9090$.
- Counting them is the same as counting the whole numbers from $910$ to $9090$, which is $9090 - 910 + 1 = 8181$.
- So there are $8181$ values of $n$ that work, which is choice (B).
💡 Counting evenly spaced numbers means counting their multiplier labels: last minus first, plus one for the endpoint.
4.NBT.B.6 Dividing $n$ by $100$ gives a quotient $q$ and a remainder $r$, which means $n = 6.EE.A.3 We care about $q + r$, but the equation shows $100q + r$. Split the $100q$ into 4.OA.B.4 The term $99q$ is always a multiple of $11$, because $99 = 9 \times 11$. So $n = 4.OA.B.4 The question is now simply: how many $5$-digit numbers are multiples of $11$? Fi 4.OA.C.5 The multiples of $11$ from $10010$ to $99990$ are $11 \times 910, 11 \times 911, Review
Reasonableness: Sanity-check the size. There are $99999 - 10000 + 1 = 90000$ five-digit numbers, and about one in every $11$ is a multiple of $11$: $90000 \div 11 \approx 8181.8$. The exact count $8181$ sits right at that estimate, so it is the right order of magnitude and it rules out the far-off choices $9000$ and $9090$. It also lands next to $8182$, so the careful endpoint count matters: the smallest qualifying $n$ is $10010$ (not $10000$) and the largest is $99990$, and $9090 - 910 + 1 = 8181$ confirms (B) over (C). A spot check of the logic: take $n = 10010$, then $q = 100$, $r = 10$, and $q + r = 110 = 11 \times 10$ is indeed divisible by $11$.
Alternative: Count directly with the pairs $(q, r)$ instead of translating to $n$. For each fixed quotient $q$ (from $100$ to $999$, that is $900$ values), we need a remainder $r$ with $0 \le r \le 99$ and $q + r$ divisible by $11$, i.e. $r$ in a fixed residue class mod $11$. Among $r = 0, 1, \ldots, 99$ (which is $100 = 9 \times 11 + 1$ values), the residue class $0$ occurs $10$ times ($0, 11, \ldots, 99$) and every other residue class occurs $9$ times. The needed class is $0$ exactly when $q$ is a multiple of $11$; the multiples of $11$ from $100$ to $999$ are $11 \times 10$ up to $11 \times 90$, i.e. $81$ values of $q$. So the total is $81 \times 10 + (900 - 81) \times 9 = 810 + 7371 = 8181$, matching the multiples-of-$11$ count. The multiples-of-$11$ route is far cleaner, which is why it is the main solution.
CCSS standards used (min grade 6)
4.NBT.B.6Find whole-number quotients and remainders with up to four-digit dividends (Turning 'quotient q and remainder r when n is divided by 100' into the equation n = 100q + r with 0 <= r <= 99.)6.EE.A.3Apply the properties of operations to generate equivalent expressions (Rewriting 100q + r as the equivalent 99q + (q + r) so the target sum q + r is exposed inside n.)4.OA.B.4Find all factor pairs and recognize multiples; determine prime or composite (Recognizing 99 = 9 x 11 as a multiple of 11, concluding 11 | (q+r) iff 11 | n, and locating the smallest and largest 5-digit multiples of 11.)4.OA.C.5Generate a number or shape pattern following a given rule (Counting the evenly spaced multiples of 11 from 11x910 to 11x9090 as 9090 - 910 + 1 = 8181.)
⭐ Since $100$ is one more than a multiple of $11$, the number $n$ and the sum $q+r$ always leave the same remainder when divided by $11$, so the question is just 'how many $5$-digit multiples of $11$ are there?' — and that count is $8181$.
⭐ Since $100$ is one more than a multiple of $11$, the number $n$ and the sum $q+r$ always leave the same remainder when divided by $11$, so the question is just 'how many $5$-digit multiples of $11$ are there?' — and that count is $8181$.
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