AMC 10 · 2002 · #3
Easy mode Grade 4Here are nine numbers: 9, 99, 999, and so on, each one having one more 9 than the last, up to 999999999 (which has nine 9s).
Add all nine numbers together, then divide the total by 9. The result is a nine-digit number M, and every digit of M is different.
Which one of these digits does M not contain?
Pick an answer.
AMC 10 2002 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 set $\{9, 99, 999, 9999, \ldots, 999999999\}$ holds nine numbers, where the $k$-th number is made of $k$ nines. Their arithmetic mean $M$ turns out to be a nine-digit number whose digits are all different. Decide which of the digits $0, 2, 4, 6, 8$ does not appear in $M$.
Givens: Nine numbers: $9$, $99$, $999$, up to $999999999$ (nine nines); The $k$-th number is a run of exactly $k$ nines; $M$ is their arithmetic mean: add all nine, then divide by $9$; $M$ is a nine-digit number and all of its digits are distinct; Answer choices for the missing digit: (A) $0$, (B) $2$, (C) $4$, (D) $6$, (E) $8$
Unknowns: The single digit from $\{0,2,4,6,8\}$ that never shows up in $M$
Understand
Restated: The set $\{9, 99, 999, 9999, \ldots, 999999999\}$ holds nine numbers, where the $k$-th number is made of $k$ nines. Their arithmetic mean $M$ turns out to be a nine-digit number whose digits are all different. Decide which of the digits $0, 2, 4, 6, 8$ does not appear in $M$.
Givens: Nine numbers: $9$, $99$, $999$, up to $999999999$ (nine nines); The $k$-th number is a run of exactly $k$ nines; $M$ is their arithmetic mean: add all nine, then divide by $9$; $M$ is a nine-digit number and all of its digits are distinct; Answer choices for the missing digit: (A) $0$, (B) $2$, (C) $4$, (D) $6$, (E) $8$
Plan
Primary tool: #5 Look for a Pattern
Secondary: #7 Identify Subproblems, #3 Eliminate Possibilities
Adding nine large numbers and long-dividing by $9$ is slow and error-prone. Instead, notice the shared structure with Tool #5 (Look for a Pattern): every number is a run of $9$s, so each equals $9$ times a run of $1$s. Tool #7 (Identify Subproblems) then splits the work — factor the $9$ out so it cancels the divide-by-$9$, then add the leftover runs of $1$s column by column. Finally Tool #3 (Eliminate Possibilities) reads off which listed digit is missing, since four of the five choices actually appear in $M$.
Execute — Answer: A
4.OA.C.5 Step 1 Read the pattern of the set
- Each number in the set is nothing but a string of $9$s, and the count grows by one each time: $9$ has one nine, $99$ has two, $999$ has three, and so on up to $999999999$ with nine nines.
- So there are exactly nine numbers, and the $k$-th one is $k$ nines in a row.
- Holding this rule in mind is what makes the shortcut possible.
💡 Spotting that the numbers are just growing runs of the same digit turns a messy list into one clean rule.
4.NBT.B.5 Step 2 Rewrite each run of 9s
- A run of $9$s is nine times the matching run of $1$s: $9 = 9\times 1$, $99 = 9\times 11$, $999 = 9\times 111$, and in general a block of $k$ nines equals $9\times(\text{a block of }k\text{ ones})$.
- This is just place value — $111\times 9 = 999$ — read backwards.
💡 A block of nines is exactly nine copies of the same block of ones.
3.OA.B.5 Step 3 Factor out the 9 and cancel
- Because every term carries a factor of $9$, pull it out of the whole sum.
- The mean divides that sum by $9$, so the factored-out $9$ cancels the divide-by-$9$ completely.
- What remains is simply the sum of the runs of $1$s — no big multiplication or long division needed.
💡 Pulling out the shared factor $9$ lets it wipe out the divide-by-$9$, so the hard arithmetic disappears.
4.NBT.B.4 Step 4 Add the runs of 1s by columns
- Stack the nine numbers $1, 11, 111, \ldots, 111111111$ and add straight down each column.
- The units column has a $1$ in all nine numbers, giving $9$.
- The tens column has a $1$ in eight of them, giving $8$.
- The hundreds column gives $7$, and so on, all the way to the far-left column, which gets a single $1$.
- No column reaches $10$, so there is no carrying, and the column totals read $1,2,3,4,5,6,7,8,9$ from left to right: $M=123456789$.
- Its digits are $1$ through $9$, all different, so the one digit it never contains is $0$.
- That is choice (A); the digits $2$, $4$, $6$, and $8$ all appear, ruling out (B), (C), (D), and (E).
💡 Each column just counts how many of the numbers are long enough to reach that place, so the totals march $9,8,7,\ldots,1$.
4.OA.C.5 Each number in the set is nothing but a string of $9$s, and the count grows by o 4.NBT.B.5 A run of $9$s is nine times the matching run of $1$s: $9 = 9\times 1$, $99 = 9\t 3.OA.B.5 Because every term carries a factor of $9$, pull it out of the whole sum. The me 4.NBT.B.4 Stack the nine numbers $1, 11, 111, \ldots, 111111111$ and add straight down eac Review
Reasonableness: Sanity-check the size: the largest number is under a billion, so the mean is under a billion too, which fits a nine-digit answer. A quick direct check confirms the shortcut — the plain sum $9+99+\cdots+999999999 = 1111111101$, and $1111111101\div 9 = 123456789$, exactly what the column method gave. Since $M=123456789$ runs through $1,2,3,4,5,6,7,8,9$, every digit except $0$ is present, so the missing one must be $0$ and all other choices are eliminated.
Alternative: Write each number as $10^k-1$: then $9=10^1-1$, $99=10^2-1$, up to $999999999=10^9-1$. The sum is $(10^1+10^2+\cdots+10^9)-9 = 1111111110-9 = 1111111101$, and dividing by $9$ gives $123456789$ — same answer, reached through powers of ten instead of runs of ones.
CCSS standards used (min grade 4)
4.OA.C.5Generate a number or shape pattern following a given rule (Recognizing that the $k$-th number in the set is a run of exactly $k$ nines.)4.NBT.B.5Multiply a whole number of up to four digits by a one-digit whole number (Rewriting each run of nines as $9$ times the matching run of ones (e.g. $999=9\times111$).)3.OA.B.5Apply properties of operations as strategies to multiply and divide (Factoring the common $9$ out of the sum so it cancels the divide-by-$9$ in the mean.)4.NBT.B.4Fluently add and subtract multi-digit whole numbers (Column-adding $1+11+\cdots+111111111$ to reach $M=123456789$ with no carrying.)
⭐ A row of nines is just nine times a row of ones, so factoring out the 9 cancels the divide-by-9 and the answer 123456789 falls out with no messy arithmetic.
⭐ A row of nines is just nine times a row of ones, so factoring out the 9 cancels the divide-by-9 and the answer 123456789 falls out with no messy arithmetic.
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