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2002 AMC 10B Problem 3

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2002 AMC 10B Problem 3AMC 10 Step by Step
Name:                             Date: September 13, 2026
Bases & Digits · difficulty 1 of 5 · about 2 min · position P1-10

The arithmetic mean of the nine numbers in the set {9,99,999,9999,…,999999999}\{9, 99, 999, 9999, \ldots, 999999999\} is a 99 -digit number MM , all of whose digits are distinct. The number MM doesn't contain the digit

  1. A)

    00

  2. B)

    22

  3. C)

    44

  4. D)

    66

  5. E)

    88

Answer

A

Key insight

Each number is 9 times a repunit, so the mean is 1 + 11 + 111 + ... + 111111111 = 123456789, which contains every digit except 0.

Solution

A string of kk nines equals 99 times a string of kk ones. So dividing each member of the set by 99 before adding, the mean is
M=1+11+111+⋯+111111111.M = 1 + 11 + 111 + \cdots + 111111111.

Add by columns. The units column contains a 11 from every one of the nine terms, so it sums to 99. The tens column gets a 11 from the eight terms with at least two digits, giving 88; the hundreds column gives 77; and so on down to the leftmost column, which gets a single 11. Every column total is at most 99, so there are no carries, and reading the columns from left to right gives
M=123456789.M = 123456789.

The digits of MM are 11 through 99; the missing digit is 00.

The answer is (A) 0\boxed{\textbf{(A)}\ 0}.

Why this works

Dividing by 99 before summing converts an ugly addition into a clean one. Repunits add column by column with no carries as long as at most nine of them overlap, so the sum spells out 1,2,3,…1, 2, 3, \ldots directly. Recognizing 99…9=9⋅11…199\ldots9 = 9 \cdot 11\ldots1 (or 10k−110^k - 1) is the standard first move with all-nines numbers.

Alternative approach

Write each term as 10k−110^k - 1. The sum is (10+102+⋯+109)−9=1111111110−9=1111111101(10 + 10^2 + \cdots + 10^9) - 9 = 1111111110 - 9 = 1111111101, and dividing by 99 by long division gives 123456789123456789.

The trap

Trying to add the nine numbers first and then divide by 9, which invites carrying errors; dividing each term by 9 first makes the column sums 1 through 9.

Common mistakes

  • Trying to add the nine numbers first and then divide by 9, which invites carrying errors; dividing each term by 9 first makes the column sums 1 through 9.
  • Reading the column totals in the wrong order and writing 987654321987654321; the digit set is the same, so the answer survives, but the reasoning is off.

Techniques

Set up the equation/formula and compute; no special trick needed · Compute small cases, spot the pattern, generalize