Divisibility and Digit Counting: Solving the Five-Digit Number Challenge

In summary, there are $\boxed{12504}$ five-digit numbers that are divisible by 3 and contain the digit 6. This is found by taking the total number of five-digit multiples of 3 (30000) and subtracting the number of five-digit multiples of 3 that do not contain the digit 6 (17496). The sum of the five digits in a five-digit multiple of 3 must be a multiple of 3, and using modulo arithmetic, we can determine that there are exactly three choices for the last digit in each case. This leads to a total of 17496 five-digit multiples of 3 without the digit 6, and subtracting this from the total number of multiples of 3 gives
  • #1
anemone
Gold Member
MHB
POTW Director
3,883
115
How many five digit numbers are divisible by 3 and contain the digit 6?
 
Mathematics news on Phys.org
  • #2
anemone said:
How many five digit numbers are divisible by 3 and contain the digit 6?
My answer is $\boxed{12504}$.
[sp]Altogether, there are $90000$ numbers with five digits. Exactly one-third of these (starting with 10002 and ending with 99999) are divisible by 3. So altogether there are $30000$ five-digit numbers divisible by 3. Now we must subtract all such numbers that contain no 6s.

To count how many five-digit multiples of 3 contain no 6s, look at how many ways there are to choose each of the five digits in the number. There are eight choices for the leading digit (which could be 1,2,3,4,5,7,8 or 9). For each of the second, third and fourth digits, there are nine choices (because we could also have a 0 in those positions). For the final digit, there are three choices. To see why, notice that the sum of the five digits must be a multiple of 3. If the sum of the first four digits is $=1\pmod3$ then the last digit must be 2,5 or 8. If the sum of the first four digits is $=2\pmod3$ then the last digit must be 1,4 or 7. And if the sum of the first four digits is $=0\pmod3$ then the last digit must be 0,3 or 9 (since it can't be 6!). So in every case there are exactly three choices.

Thus altogether there are $8*9^3*3 = 17496$ five-digit multiples of 3 containing no 6s, and subtracting that number from $30000$ we get $12504$ as the answer.[/sp]
 
  • #3
Well done, Opalg! Your answer is of course correct and for your information, I didn't solve this problem, however, I did find a solution which was very nearly identical to yours, and I never realized that modulo arithmetic could be a tool to solve counting problems like this.

I like your explanation so much and whenever it has come time to read your explanatory posts, you have never failed to enlighten us with your clever step-by-step working and so I thank you for participating and for this brilliant solution! My hat is truly off to you, Mr. Opalg!
 

FAQ: Divisibility and Digit Counting: Solving the Five-Digit Number Challenge

What is the "Counting Techniques Challenge"?

The "Counting Techniques Challenge" is a mathematical problem-solving activity that involves using various counting techniques to solve a given problem. It is often used as a tool for developing critical thinking and problem-solving skills.

What are some examples of counting techniques used in the challenge?

Some examples of counting techniques used in the challenge include combinations, permutations, and the inclusion-exclusion principle. These techniques are used to count the number of possible outcomes in a given scenario.

Who can participate in the "Counting Techniques Challenge"?

The "Counting Techniques Challenge" is open to anyone with an interest in mathematics and problem-solving. It is often used in academic settings, but can also be enjoyed by individuals or groups as a fun and challenging activity.

What are the benefits of participating in the "Counting Techniques Challenge"?

The "Counting Techniques Challenge" can help develop critical thinking and problem-solving skills, as well as improve overall mathematical ability. It can also be a fun and engaging way to learn about different counting techniques and their applications.

Are there any resources available to help with the "Counting Techniques Challenge"?

Yes, there are many resources available online, such as practice problems, tutorials, and video explanations, to help individuals or groups with the "Counting Techniques Challenge." Additionally, many textbooks and mathematics websites also provide information and examples of counting techniques.

Similar threads

Replies
1
Views
751
Replies
3
Views
1K
Replies
11
Views
1K
Replies
7
Views
1K
Replies
20
Views
2K
Replies
3
Views
1K
Replies
1
Views
2K
Back
Top