Math Problem Statement

How many 4-digit numbers can be formed if the number must be divisible by 2 and digits cannot be repeated?

Solution

To solve the problem of how many 4-digit numbers can be formed under the given conditions—divisible by 2 and with no repeated digits—we can break it down into steps:

Conditions:

  1. The number must be divisible by 2. This means the last digit must be an even number (0, 2, 4, 6, or 8).
  2. The digits cannot be repeated.

Step-by-step approach:

Step 1: Determine possible values for the last digit

Since the number must be divisible by 2, the last digit has to be one of the even digits: {0, 2, 4, 6, 8}. Therefore, there are 5 possible choices for the last digit.

Step 2: Determine possible values for the first digit

The first digit cannot be 0 (since the number must be a 4-digit number), and it also cannot be the same as the last digit (to satisfy the condition of no repetition of digits). So, for the first digit, we can choose from the remaining 8 digits (1–9 excluding the last digit).

Step 3: Determine possible values for the second digit

Once the first digit is chosen, we have 8 remaining digits (since one digit is used for the first digit and one for the last digit). So, we have 8 possible choices for the second digit.

Step 4: Determine possible values for the third digit

At this point, two digits have already been used (the first and second digits), so we have 7 digits left to choose from for the third digit.

Step 5: Calculate the total number of possible combinations

Using the choices from the steps above:

  • First digit: 8 choices
  • Second digit: 8 choices
  • Third digit: 7 choices
  • Last digit: 5 choices (since it must be even)

Thus, the total number of 4-digit numbers is: 8×8×7×5=22408 \times 8 \times 7 \times 5 = 2240

Answer:

2240 different 4-digit numbers can be formed that are divisible by 2 with no repeated digits.

Would you like any further details or clarification?

Here are some related questions:

  1. How would the solution change if repeated digits were allowed?
  2. What if the number needed to be divisible by 5 instead of 2?
  3. How many 3-digit numbers can be formed under the same conditions?
  4. What if the number needed to be divisible by 3 and not 2?
  5. How many 4-digit even numbers can be formed if 0 cannot be used at all?

Tip: When solving problems involving divisibility and constraints, always consider constraints step by step to avoid missing details.

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Math Problem Analysis

Mathematical Concepts

Number Theory
Combinatorics
Divisibility Rules

Formulas

Total combinations = choices for first digit × choices for second digit × choices for third digit × choices for last digit

Theorems

Divisibility rule for 2

Suitable Grade Level

Grades 9-11