How many three-digit positive integers have an odd number of even digits?
- A)
- B)
- C)
- D)
- E)
Answer
D
Key insight
Fix the first two digits; among the ten possible units digits exactly five flip the parity of the even-digit count, so exactly half of all 900 numbers work.
Solution
Choose the hundreds and tens digits first: ways. Those two digits contain some number of even digits, either even or odd.
Now the units digit runs over : five even values and five odd values. If the first two digits already contain an odd number of even digits, the five odd units digits keep the count odd; if they contain an even number, the five even units digits make it odd. Either way exactly of the units digits work.
Total: .
The answer is .
Why this works
The last digit acts as a "parity switch": whatever the earlier digits did, half of the choices for the final digit produce each parity. This pairing argument shows that exactly half of the three-digit numbers qualify, with no casework at all. Look for a free position that can flip the property you care about.
Alternative approach
Direct casework. Exactly one even digit: even hundreds digit with two odd digits, ; odd hundreds digit with exactly one of the other two even, . Exactly three even digits: . Sum .
The trap
Treating the hundreds digit like the others (it cannot be 0, so only 4 of its 9 values are even) when doing explicit casework.
Common mistakes
- Treating the hundreds digit like the others (it cannot be 0, so only 4 of its 9 values are even) when doing explicit casework.
- Forgetting the "three even digits" case and answering 350.
Techniques
Exploit symmetry to reduce work or pair up objects