Difference between revisions of "2022 AMC 10B Problems/Problem 3"
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==Solution 1== | ==Solution 1== | ||
| − | + | We use simple case work to solve this problem. | |
Case 1: even, even, even = <math>4 \cdot 5 \cdot 5 = 100</math> | Case 1: even, even, even = <math>4 \cdot 5 \cdot 5 = 100</math> | ||
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Case 4: odd, odd, even = <math>5 \cdot 5 \cdot 5 = 125</math> | Case 4: odd, odd, even = <math>5 \cdot 5 \cdot 5 = 125</math> | ||
| − | + | Simply sum up the cases to get your answer, <math>100 + 100 + 125 + 125 = \boxed{\textbf{(D)~}450}</math>. | |
- Wesseywes7254 | - Wesseywes7254 | ||
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~mathboy100 | ~mathboy100 | ||
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==Video Solution 1 (🚀Just 2 min🚀)== | ==Video Solution 1 (🚀Just 2 min🚀)== | ||
Latest revision as of 19:05, 29 October 2024
Contents
Problem
How many three-digit positive integers have an odd number of even digits?
Solution 1
We use simple case work to solve this problem.
Case 1: even, even, even = 
Case 2: even, odd, odd =
Case 3: odd, even, odd = 
Case 4: odd, odd, even =
Simply sum up the cases to get your answer, 
.
- Wesseywes7254
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 (🚀Just 2 min🚀)
~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. 
