How Many 3-Digit Even Numbers Can Be Formed?

In this post, we’ll explore how to count the number of 3-digit even numbers that can be formed. We will break down the problem step by step using simple language, so that anyone can understand the process.

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Introduction

A 3-digit number ranges from 100 to 999. For a number to be even, its last digit must be one of the even digits: 0, 2, 4, 6, or 8. Our goal is to figure out how many numbers within this range meet the “even” condition. We will do this by looking at the choices for each digit in a 3-digit number.

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Breaking Down the Problem

A 3-digit number has three positions: the hundreds place, the tens place, and the units (ones) place. Each position has a set of choices:

1. Hundreds Place:
   • The hundreds digit cannot be 0 (or the number would not be a 3-digit number).
   • Choices for the hundreds digit: 1 through 9, which gives us 9 possibilities.
2. Tens Place:
   • The tens digit can be any digit from 0 to 9.
   • This gives us 10 possibilities.
3. Units Place (Ones Place):
   • Since the number must be even, the units digit must be one of the even digits: 0, 2, 4, 6, or 8.
   • This gives us 5 possibilities.

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Calculating the Total

To find the total number of 3-digit even numbers, we multiply the number of choices for each digit: Total 3-digit even numbers=(Choices for hundreds)×(Choices for tens)×(Choices for units)\text{Total 3-digit even numbers} = (\text{Choices for hundreds}) \times (\text{Choices for tens}) \times (\text{Choices for units})

Substituting the numbers: Total=9×10×5=450\text{Total} = 9 \times 10 \times 5 = 450

So, there are 450 3-digit even numbers.

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Detailed Explanation

• Hundreds Digit (9 choices):
  The hundreds digit is the first digit of a 3-digit number and must be between 1 and 9. Zero is not allowed here because it would make the number less than 100.
• Tens Digit (10 choices):
  The tens digit has no restrictions—it can be any digit from 0 to 9. This includes 0, which is acceptable in the middle of a number.
• Units Digit (5 choices):
  The units digit determines whether the number is even. For a number to be even, its last digit must be 0, 2, 4, 6, or 8. Hence, there are 5 choices for this digit.

When these choices are combined, each independent choice multiplies with the others. This is why we multiply 9, 10, and 5 together to get 450.

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Conclusion

By following the steps above, we can clearly see how many 3-digit even numbers can be formed:

• Hundreds digit: 9 options
• Tens digit: 10 options
• Units digit (even only): 5 options

Multiplying these together gives: 9×10×5=4509 \times 10 \times 5 = 450

Thus, there are 450 3-digit even numbers. This method of breaking the problem down into smaller parts can be applied to many other counting problems as well. Happy calculating!

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