2022 AMC 10B Problem 3

Attempt Problem 3 of the 2022 AMC 10B below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2022 AMC 10B solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

How many three-digit positive integers have an odd number of even digits?

$150$

$250$

$350$

$450$

$550$

Answer: D

Concepts:digits parity multiplication principle

Difficulty rating: 1100

Solution:

There are $9\cdot10=90$ choices for the hundreds and tens digits. Once those two digits are fixed, the units digit must have whichever parity makes the total number of even digits odd. There are always $5$ digits of the required parity (including $0$ among the even digits).

Therefore, the number of integers is $90\cdot5=450.$

Thus, the answer is D .

← Problem 2#2

Full Exam

Problem 4#4 →

Problem 3 in Other Years

2000 AMC 10 · 2001 AMC 10 · 2002 AMC 10A · 2002 AMC 10B · 2003 AMC 10A · 2003 AMC 10B · 2004 AMC 10A · 2004 AMC 10B · 2005 AMC 10A · 2005 AMC 10B · 2006 AMC 10A · 2006 AMC 10B · 2007 AMC 10A · 2007 AMC 10B · 2008 AMC 10A · 2008 AMC 10B · 2009 AMC 10A · 2009 AMC 10B · 2010 AMC 10A · 2010 AMC 10B · 2011 AMC 10A · 2011 AMC 10B · 2012 AMC 10A · 2012 AMC 10B · 2013 AMC 10A · 2013 AMC 10B · 2014 AMC 10A · 2014 AMC 10B · 2015 AMC 10A · 2015 AMC 10B · 2016 AMC 10A · 2016 AMC 10B · 2017 AMC 10A · 2017 AMC 10B · 2018 AMC 10A · 2018 AMC 10B · 2019 AMC 10A · 2019 AMC 10B · 2020 AMC 10A · 2020 AMC 10B · 2021 AMC 10A Spring · 2021 AMC 10B Spring · 2021 AMC 10A Fall · 2021 AMC 10B Fall · 2022 AMC 10A · 2023 AMC 10A · 2023 AMC 10B · 2024 AMC 10A · 2024 AMC 10B · 2025 AMC 10A · 2025 AMC 10B