1993 AMC 8 Problem 22

Below is the professionally curated solution for Problem 22 of the 1993 AMC 8, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1993 AMC 8 solutions, or check the answer key.

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Concepts:digitssystematic listing

Difficulty rating: 1200

22.

Pat Peano has plenty of 00's, 11's, 33's, 44's, 55's, 66's, 77's, 88's and 99's, but he has only twenty-two 22's. How far can he number the pages of his scrapbook with these digits?

2222

9999

112112

119119

199199

Solution:

Numbering 11 through 9999 uses ten 22's in the units place and ten in the tens place, a total of twenty 22's. Pages 100100 and 101101 use none.

The remaining two 22's are used on pages 102102 and 112.112. After that, pages 113113 through 119119 need no 2,2, but 120120 would require another 2,2, so he can number up to 119.119.

Thus, the correct answer is D .

Problem 22 in Other Years

1985 AMC 8 · 1986 AMC 8 · 1987 AMC 8 · 1988 AMC 8 · 1989 AMC 8 · 1990 AMC 8 · 1991 AMC 8 · 1992 AMC 8 · 1994 AMC 8 · 1995 AMC 8 · 1996 AMC 8 · 1997 AMC 8 · 1998 AMC 8 · 1999 AMC 8 · 2000 AMC 8 · 2001 AMC 8 · 2002 AMC 8 · 2003 AMC 8 · 2004 AMC 8 · 2005 AMC 8 · 2006 AMC 8 · 2007 AMC 8 · 2008 AMC 8 · 2009 AMC 8 · 2010 AMC 8 · 2011 AMC 8 · 2012 AMC 8 · 2013 AMC 8 · 2014 AMC 8 · 2015 AMC 8 · 2016 AMC 8 · 2017 AMC 8 · 2018 AMC 8 · 2019 AMC 8 · 2020 AMC 8 · 2022 AMC 8 · 2023 AMC 8 · 2024 AMC 8 · 2025 AMC 8 · 2026 AMC 8