Math Problem Statement
How many four-digit numbers, which are divisible by 6, can be formed using the digits 0, 2, 3, 4, 6, such that no digit is used more than once and 0 does not occur in the left-most position?
Solution
To solve this problem, we need to calculate how many four-digit numbers divisible by 6 can be formed using the digits , where no digit is repeated and is not the first digit.
A number is divisible by 6 if it satisfies two conditions:
- Divisibility by 2: The last digit must be even.
- Divisibility by 3: The sum of the digits must be divisible by 3.
Let's go step by step:
Step 1: Available digits
The digits we can use are , and there are five available digits. We are tasked with creating a four-digit number, so we will select four of these digits.
Step 2: Condition for divisibility by 2
Since the number must be divisible by 2, the last digit must be an even digit. The available even digits are . We will select the last digit from this set.
Step 3: Condition for divisibility by 3
For divisibility by 3, the sum of the digits of the number must be divisible by 3. The sum of all available digits () is: Since 15 is divisible by 3, removing any digit will still result in a sum divisible by 3, meaning we don’t need to worry about this condition—it will automatically be satisfied.
Step 4: Selecting the last digit
Now, let’s calculate the number of possible four-digit numbers based on each case for the last digit:
-
Case 1: Last digit is 0
The first digit cannot be 0, and we have left to choose from for the first three digits.- The first digit can be chosen from {2, 3, 4, 6}, so there are 4 choices.
- The second digit can be chosen from the remaining 3 digits.
- The third digit can be chosen from the remaining 2 digits. So, the total number of possibilities in this case is:
-
Case 2: Last digit is 2
The remaining digits are {0, 3, 4, 6}.- The first digit cannot be 0, so it must be chosen from {3, 4, 6}, giving 3 choices.
- The second digit can be chosen from the remaining 3 digits.
- The third digit can be chosen from the remaining 2 digits. So, the total number of possibilities in this case is:
-
Case 3: Last digit is 4
The remaining digits are {0, 2, 3, 6}.- The first digit cannot be 0, so it must be chosen from {2, 3, 6}, giving 3 choices.
- The second digit can be chosen from the remaining 3 digits.
- The third digit can be chosen from the remaining 2 digits. So, the total number of possibilities in this case is:
-
Case 4: Last digit is 6
The remaining digits are {0, 2, 3, 4}.- The first digit cannot be 0, so it must be chosen from {2, 3, 4}, giving 3 choices.
- The second digit can be chosen from the remaining 3 digits.
- The third digit can be chosen from the remaining 2 digits. So, the total number of possibilities in this case is:
Step 5: Total possibilities
Summing the possibilities from all four cases:
Thus, the total number of four-digit numbers divisible by 6 that can be formed using the digits without repetition and with no in the left-most position is .
Would you like a detailed explanation for any specific part?
Here are five related questions:
- How would the solution change if repetition of digits was allowed?
- What if the number had to be divisible by 5 instead of 6?
- How many three-digit numbers divisible by 6 can be formed using the same set of digits?
- How would you handle the case where the first digit can be zero?
- What if you were allowed to use five digits instead of four?
Tip: Always check divisibility rules carefully, as they simplify the process of eliminating invalid numbers.
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Math Problem Analysis
Mathematical Concepts
Number Theory
Divisibility Rules
Permutations
Formulas
Divisibility by 6: last digit divisible by 2, sum of digits divisible by 3
Theorems
Divisibility Rule for 2
Divisibility Rule for 3
Suitable Grade Level
Grades 9-12
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