# 2022 AMC 10B Problems/Problem 3

## 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 (Casework)

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

Case 1: $\bf{1}$ 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: $\bf{3}$ 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

## 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🚀)

~Education, the Study of Everything

## Video Solution by Interstigation

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