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Difference between revisions of "2022 AMC 10B Problems/Problem 3"

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<math>\textbf{(A) }150\qquad\textbf{(B) }250\qquad\textbf{(C) }350\qquad\textbf{(D) }450\qquad\textbf{(E) }550</math>
 
<math>\textbf{(A) }150\qquad\textbf{(B) }250\qquad\textbf{(C) }350\qquad\textbf{(D) }450\qquad\textbf{(E) }550</math>
  
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==Solution 1 (Casework) ==
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==Solution 1==
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We use simple case work to solve this problem.
  
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There are only <math>2</math> ways for an odd number of even digits: <math>1</math> even digit or all even digits.
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Case 1: even, even, even = <math>4 \cdot 5 \cdot 5 = 100</math>
  
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'''Case 1: <math>\bf{1}</math> even digit'''
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Case 2: even, odd, odd = <math>4 \cdot 5 \cdot 5 = 100</math>
  
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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.
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Case 3: odd, even, odd = <math>5 \cdot 5 \cdot 5 = 125</math>
  
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'''Case 2: all even digits'''
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Case 4: odd, odd, even = <math>5 \cdot 5 \cdot 5 = 125</math>
  
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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>.
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Simply sum up the cases to get your answer, <math>100 + 100 + 125 + 125 = \boxed{\textbf{(D)~}450}</math>.
  
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Adding up the cases, the answer is <math>100+350=\boxed{\textbf{(D) }450}</math>.
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- Wesseywes7254
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~MrThinker
 
  
 
==Solution 2 (Bijection)==
 
==Solution 2 (Bijection)==
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~mathboy100
 
~mathboy100
  
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==Video Solution 1==
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==Video Solution 1 (🚀Just 2 min🚀)==
 
https://youtu.be/rAad-1GMgIs
 
https://youtu.be/rAad-1GMgIs
  
 
~Education, the Study of Everything
 
~Education, the Study of Everything
 +
 
==Video Solution by Interstigation==
 
==Video Solution by Interstigation==
 
https://youtu.be/_KNR0JV5rdI?t=167
 
https://youtu.be/_KNR0JV5rdI?t=167

Latest revision as of 19:05, 29 October 2024

Problem

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

$\textbf{(A) }150\qquad\textbf{(B) }250\qquad\textbf{(C) }350\qquad\textbf{(D) }450\qquad\textbf{(E) }550$

Solution 1

We use simple case work to solve this problem.

Case 1: even, even, even = $4 \cdot 5 \cdot 5 = 100$

Case 2: even, odd, odd = $4 \cdot 5 \cdot 5 = 100$

Case 3: odd, even, odd = $5 \cdot 5 \cdot 5 = 125$

Case 4: odd, odd, even = $5 \cdot 5 \cdot 5 = 125$

Simply sum up the cases to get your answer, $100 + 100 + 125 + 125 = \boxed{\textbf{(D)~}450}$.

- Wesseywes7254

Solution 2 (Bijection)

We will show that the answer is $450$ 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 $1$, unless the last digit is $9$, in which case it becomes $0$. 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 $\boxed{\textbf{(D)~}450}$

~mathboy100


Video Solution 1 (🚀Just 2 min🚀)

https://youtu.be/rAad-1GMgIs

~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

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