Problem 37
Question
How many numbers can be made with the digits \(3,4,5,6,7,8\) lying between 3000 and 4000, which are divisible by 5 while repetition of any digit is not allowed in any number? a. 60 b. 12 c. 120 d. 24
Step-by-Step Solution
Verified Answer
b. 12
1Step 1: Understanding the Criteria
We need to form a number using the provided digits that lies between 3000 and 4000 and is divisible by 5. Additionally, we cannot repeat any digit.
2Step 2: Determining the Thousands Place
To ensure the number is between 3000 and 4000, the thousands digit must be 3. Thus, the number will be in the form of \(3xyz\), where \(x\), \(y\), and \(z\) are other digits selected from the remaining digits.
3Step 3: Ensuring Divisibility by 5
For a number to be divisible by 5, the units digit (\(z\)) must be either 0 or 5. Since our available digits are 3, 4, 5, 6, 7, and 8, the units digit must be 5.
4Step 4: Constructing the Number
With the thousands place being 3 and the units place being 5, the number takes the form \(3xy5\). We need to determine the suitable values for \(x\) and \(y\) using the digits 4, 6, 7, and 8.
5Step 5: Calculating Combinations for Hundreds and Tens
There are 4 digits (4, 6, 7, and 8) left for digits \(x\) and \(y\). Choose any 2 digits for \(x\) in 4 ways, and the remaining digit for \(y\) in 3 ways. Thus, the total numbers can be calculated as \(4 \times 3 = 12\).
6Step 6: Final Calculation and Verification
Verify whether the selection satisfies all the conditions: 12 numbers are formed which satisfy that they lie between 3000 and 4000 and are divisible by 5 without repeating any digit.
Key Concepts
Number DivisibilityDigit ArrangementConstraints in Number Formation
Number Divisibility
Divisibility is a concept in mathematics where one number can be divided by another without leaving a remainder. In our exercise, we are particularly concerned with divisibility by 5.
For a number to be divisible by 5, its last digit must be either 0 or 5.
This is a key rule derived from the basic fact that any multiple of 5 will end with these digits.
For a number to be divisible by 5, its last digit must be either 0 or 5.
This is a key rule derived from the basic fact that any multiple of 5 will end with these digits.
- In this problem, the digits we can use are 3, 4, 5, 6, 7, and 8.
- Since these do not contain 0, the digit 5 is our only choice to ensure the number meets the divisibility rule.
Digit Arrangement
Arranging digits to form numbers is central to combinatorics. It involves choosing which digits appear in which positions to fulfill certain criteria.
In this exercise, we arrange the digits 3, 4, 5, 6, 7, and 8.
Let's break down the arrangement process:
This method of sequentially filling digit positions ensures each number fits the stated requirements.
In this exercise, we arrange the digits 3, 4, 5, 6, 7, and 8.
Let's break down the arrangement process:
- The thousands place must be 3 to satisfy the range requirement.
- The units place must be 5 to ensure divisibility by 5, as discussed earlier.
- Once these positions are fixed, the remaining digits (4, 6, 7, 8) are available for the hundreds and tens places.
This method of sequentially filling digit positions ensures each number fits the stated requirements.
Constraints in Number Formation
Constraints are the rules that restrict the choice of digit in specific places when forming numbers. In this problem, constraints play a significant part in guiding the formation process.
Here's how constraints influence our solution:
This problem reduces to arranging the four remaining digits after fixing the thousands and units digits. Adhering to these constraints leads to a controlled and structured approach in number formation, allowing us to precisely determine that there are 12 viable combinations fitting all criteria.
Here's how constraints influence our solution:
- The number must be between 3000 and 4000. This means the thousands digit is fixed as 3.
- The divisibility by 5 condition determines the units digit to be 5.
- Since no digit can be repeated and only specific digits are available, every choice for hundreds or tens affects subsequent selections.
This problem reduces to arranging the four remaining digits after fixing the thousands and units digits. Adhering to these constraints leads to a controlled and structured approach in number formation, allowing us to precisely determine that there are 12 viable combinations fitting all criteria.
Other exercises in this chapter
Problem 35
The number of ordered pairs of integers \((x, y)\) satisfying the equation \(x^{2}+6 x+y^{2}=4\) is a. 2 b. 8 c. 6 d. none of these
View solution Problem 36
The number of five-digit telephone numbers having at least one of their digits repeated is a. 90000 b. \(\quad 100000\) c. 30240 d. 69760
View solution Problem 38
The number of ways in which ten candidates \(A_{1}, A_{2}, \ldots, A_{10}\) can be ranked such that \(A_{1}\) is always above \(A_{10}\) is a. \(5 !\) b. \(2(5
View solution Problem 39
In the decimal system of numeration of six-digit numbers in which the sum of the digits is divisible by 5 is a. 180000 b. 540000 c. \(5 \times 10^{5}\) d. none
View solution