Difference between revisions of "2022 AMC 10B Problems/Problem 3"
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There are <math>5 \cdot 5 = 25</math> ways to choose the odd digits, <math>5</math> ways for the even digit, and <math>3</math> ways to order the even digit. So, <math>25 \cdot 5 \cdot 3 = 375</math>. However, there are <math>5 \cdot 5= 25</math> ways that the hundred's digit is <math>0</math> and we must subtract this from <math>375</math>, leaving us with <math>350</math> ways. | There are <math>5 \cdot 5 = 25</math> ways to choose the odd digits, <math>5</math> ways for the even digit, and <math>3</math> ways to order the even digit. So, <math>25 \cdot 5 \cdot 3 = 375</math>. However, there are <math>5 \cdot 5= 25</math> ways that the hundred's digit is <math>0</math> and we must subtract this from <math>375</math>, leaving us with <math>350</math> ways. | ||
| − | '''Case 2: | + | '''Case 2: <math>\bf{3}</math> even digits''' |
There are <math>5 \cdot 5 \cdot 5 = 125</math> ways to choose the even digits, and <math>5 \cdot 5 = 25</math> ways where the hundred's digit is <math>0</math>. So, <math>125-25=100</math>. | There are <math>5 \cdot 5 \cdot 5 = 125</math> ways to choose the even digits, and <math>5 \cdot 5 = 25</math> ways where the hundred's digit is <math>0</math>. So, <math>125-25=100</math>. | ||
Revision as of 20:20, 27 July 2023
Contents
Problem
How many three-digit positive integers have an odd number of even digits?
Solution 1 (Casework)
There are only 
ways for an odd number of even digits: 
even digit or all even digits.
Case 1: 
even digit
There are 
ways to choose the odd digits, 
ways for the even digit, and 
ways to order the even digit. So, 
. However, there are 
ways that the hundred's digit is 
and we must subtract this from 
, leaving us with 
ways.
Case 2: 
even digits
There are 
ways to choose the even digits, and ways where the hundred's digit is . So, 
.
Adding up the cases, the answer is 
.
~MrThinker
Solution 2 (Bijection)
We will show that the answer is 
by proving a bijection between the three digit integers that have an even number of even digits and the three digit integers that have an odd number of even digits. For every even number with an odd number of even digits, we increment the number's last digit by , unless the last digit is 
, in which case it becomes . It is very easy to show that every number with an even number of even digits is mapped to every number with an odd number of even digits, and vice versa. Thus, the answer is half the number of three digit numbers, or 
~mathboy100
Video Solution 1
~Education, the Study of Everything
Video Solution by Interstigation
https://youtu.be/_KNR0JV5rdI?t=167
See Also
| 2022 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 2 |
Followed by Problem 4 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| All AMC 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
