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

(Created page with "==Problem== How many three-digit positive integers have an odd number of even digits? <math>\textbf{(A) }150\qquad\textbf{(B) }250\qquad\textbf{(C) }350\qquad\textbf{(D) }45...")
 
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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.  
 
There are only <math>2</math> ways for an odd number of even digits: <math>1</math> even digit or all even digits.  
  
−
'''Case 1''': <math>1</math> even digit
+
'''Case 1: <math>1</math> even digit'''
  
 
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''': all even digits
+
'''Case 2: all 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 14:06, 17 November 2022

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

There are only $2$ ways for an odd number of even digits: $1$ even digit or all even digits.

Case 1: $1$ even digit

There are $5 \cdot 5 = 25$ ways to choose the odd digits, $5$ ways for the even digit, and $3$ ways to order the even digit. So, $25 \cdot 5 \cdot 3 = 375$. However, there are $5 \cdot 5= 25$ ways that the hundred's digit is $0$ and we must subtract this from $375$, leaving us with $350$ ways.

Case 2: all even digits

There are $5 \cdot 5 \cdot 5 = 125$ ways to choose the even digits, and $5 \cdot 5 = 25$ ways where the hundred's digit is $0$. So, $125-25=100$.

Adding up the cases, the answer is $100+350=\boxed{\textbf{(D) }450}$.

~MrThinker

See Also

2022 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 3 		Followed by
Problem 5
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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