Question

Answer the following questions about 9 -digit natural numbers with no repeated digits (leading zeros are not permitted): (a) How many such 9-digit numbers exist? (b) How many are divisible by \(2 ?\) (c) How many are divisible by \(5 ?\) (d) How many are greater than \(500,000.000 ?\)

Short answer

(a) 362,880; (b) 161,280; (c) 40,320; (d) 201,600.

Step-by-step solution

Counting 9-digit numbers with no repeated digits

First, calculate how many 9-digit numbers can be formed using the digits 1 to 9. Each digit must be unique, and leading zeros aren't permitted:- Choose 9 different digits from 1 to 9.- The order of these digits matters as we are forming a number. - This is a permutation of 9 digits, which can be calculated as: \[ 9! = 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 362,880. \] Thus, there are 362,880 such 9-digit numbers.

Counting numbers divisible by 2

To determine how many of these numbers are divisible by 2, the last digit must be even, i.e., one of 2, 4, 6, or 8. - We have 4 options for the last digit.- After selecting the last digit, the remaining 8 digits can be any of the 8 remaining digits. The number of ways to arrange these is: \[ 8! = 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 40,320. \]- Therefore, there are \[4 \times 8! = 4 \times 40,320 = 161,280 \] such numbers.

Counting numbers divisible by 5

To be divisible by 5, the last digit must be 5, as it is the only allowable 5-digit.- Choose 5 as the last digit (1 option).- The remaining 8 digits can again be any of 8 remaining options, which means the number of permutations of these 8 digits is: \[ 8! = 40,320. \]- Therefore, there are 40,320 such numbers.

Counting numbers greater than 500,000,000

To be greater than 500,000,000, the leading digit must be either 5, 6, 7, 8, or 9. - There are 5 options for the first digit.- The remaining 8 digits can be any of the remaining 8, fully permuted, yielding: \[ 8! = 40,320. \]- Therefore, there are \[5 \times 8! = 5 \times 40,320 = 201,600 \] such numbers.

Key concepts

9-digit numbers

When considering 9-digit numbers, it's essential to understand that we're dealing with a specific range of numbers composed only of the digits 1 through 9.
Each digit is used exactly once, which means repeating digits are not allowed. This restriction brings a level of complexity and intrigue to creating such numbers.
For example, a valid 9-digit number in this context might be 123456789, while 112345678 would not be valid as the digit '1' repeats. The exclusion of 0 further simplifies calculations, ensuring that each number remains a substantial 9 digits long.
Understanding 9-digit numbers is crucial for solving specific problems involving permutations and divisibility, as seen in exercises involving large number calculations.

Divisibility

Divisibility is a fundamental mathematical concept where one number can be divided by another without leaving a remainder. In the context of 9-digit numbers, checking divisibility is crucial, especially when trying to find numbers divisible by specific digits like 2, 5, or others.
For a number to be divisible by 2, its last digit must be an even number, such as 2, 4, 6, or 8. Conversely, for divisibility by 5, the number's last digit must be 0 or 5; but considering our constraint of using digits 1 through 9, only 5 works.
Understanding these basic rules of divisibility helps you quickly deduce how many numbers fit the criteria, streamlining the problem-solving process significantly.

Natural numbers

Natural numbers are a fundamental concept in mathematics, used to count and order things. They are the set of positive integers starting from 1, meaning we use numbers like 1, 2, 3, and so on. In our specific discussion, we are considering 9-digit natural numbers constructed using each digit from 1 to 9 exactly once.
This means every number involved in our exercise has significant place value while still adhering to the properties of natural numbers: no fractions, negatives, or decimals.
Utilizing natural numbers in exercises helps maintain a straightforward approach, allowing us to focus on the simplicity and purity of calculation and arrangement necessary for permutation problems.

Permutation calculations

Permutation calculations play a crucial role when figuring out the number of possible arrangements for a given set of items. With 9-digit numbers where each digit is unique, permutation calculations help us determine how many different ways we can arrange these numbers.
The main formula used here is the factorial notation, represented as \(n!\), which translates to multiplying a series of descending natural numbers. In our exercise, forming a 9-digit number involves calculating \(9!\), equating to \[9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 362,880.\]
This factorial calculation gives the total number of ways to arrange 9 distinct digits, and understanding this concept is key to solving permutation problems efficiently.