AMC 10 · 2002 · #3

Grade 4 arithmetic
mean-median-mode-rangeplace-valuemulti-digit-arithmetic pattern-recognition ↑ Prerequisites: mean-median-mode-range
📏 Medium solution 💡 2 insights
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Problem
The set {9, 99, 999, 9999, …, 999999999} holds nine numbers, where the k-th number is made of k nines. Their arithmetic mean M is a nine-digit number whose digits are all different. Which of the digits 0, 2, 4, 6, 8 does not appear in M?

Pick an answer.

(A)
$\ 0$
(B)
$\ 2$
(C)
$\ 4$
(D)
$\ 6$
(E)
$\ 8$

AMC 10 2002 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Look for a Pattern

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 9s, so each equals 9 times a run of 1s. 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 1s 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.

1STEP 1

Read the pattern of the set

Every number is a run of 9s, one nine longer each time: nine numbers in all, the k-th being k nines.

9₁, 99₂, 999₃, …, 999999999₉
2STEP 2

Rewrite each run of 9s

A run of 9s is nine times the matching run of 1s: 9 = 9×1, 99 = 9×11, 999 = 9×111.

9=9·1, 99=9·11, 999=9·111, …, 999999999=9·111111111
3STEP 3

Factor out the 9 and cancel

Pull the shared 9 out of the sum; it cancels the divide-by-9, leaving only 1 + 11 + … + 111111111.

M=(9 · 1+9 · 11+…+9 · 111111111)/9=(9 (1+11+…+111111111))/9=1+11+…+111111111
4STEP 4

Add the runs of 1s by columns

Stack them and add down: the columns give 9, 8, 7, …, 1 with no carrying, so M = 123456789.

1+11+111+…+111111111=123456789 → (A) 0
Answer
0
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+…+999999999 = 1111111101, and 1111111101 ÷ 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.
💡Key takeaway

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.

  • Read the pattern of the set
  • Rewrite each run of 9s
  • Factor out the 9 and cancel
  • Add the runs of 1s by columns