The number of numbers greater than 5000, less than 9000 and divisible by 3, that can be formed using

Q: Wait, why is the first digit fixed at 5? Can't it be 6, 7, or 8?

That is a great observation! Look closely at the set of available digits: $\{0, 1, 2, 5, 9\}$. The range is strictly between $5000$ and $9000$. If the first digit were $0, 1,$ or $2$, the number would be too small. If it were $9$, the number would be at least $9000$, which violates the upper bound. Thus, $5$ is the only digit from our set that keeps the number within the valid range.

Q: Why do we use modular arithmetic? Can't we just list them out?

Listing is a trap! In JEE Advanced, time is your most precious currency. Modular arithmetic transforms a tedious counting problem into a structured algebraic one. It prevents errors and ensures you do not miss any cases. By focusing on the remainders, we reduce the complexity of the problem significantly.

Q: Why do we arrange the remainders using $\frac{3!}{2!}$?

That is the permutation formula for items with repetition. Since we have two identical remainders in each case (e.g., $0, 0, 1$), we must divide by the factorial of the count of repeating items to avoid overcounting. It is the standard way to handle permutations when some items are indistinguishable.

Question

The number of numbers greater than 5000, less than 9000 and divisible by 3, that can be formed using the digits 0, 1, 2, 5, 9, if the repetition of the digits is allowed, is ......... .

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We are tasked with finding the number of integers $N$ such that $5000 < N < 9000$, where $N$ is divisible by $3$, using the digits $\{0, 1, 2, 5, 9\}$ with repetition allowed.

Since $5000 < N < 9000$, the thousands digit must be $5$. Any digit smaller than $5$ results in a number less than $5000$, and any digit larger than $5$ (such as $9$) results in a number greater than $9000$.

Thus, the number is of the form $5 d_2 d_3 d_4$. We have successfully reduced the problem to determining the number of ways to choose the three remaining digits.

### The Master Equation

A number is divisible by $3$ if and only if the sum of its digits is a multiple of $3$. Mathematically, this is expressed as:

$$(5 + d_2 + d_3 + d_4) \equiv 0 \pmod 3$$

Since $5 \equiv 2 \pmod 3$, the condition simplifies to:

$$(2 + d_2 + d_3 + d_4) \equiv 0 \pmod 3 \implies (d_2 + d_3 + d_4) \equiv 1 \pmod 3$$

This is our master condition. We must choose three digits from the set $\{0, 1, 2, 5, 9\}$ such that their sum leaves a remainder of $1$ when divided by $3$.

### The Categorization Strategy

To solve this systematically, we categorize the available digits by their remainders modulo $3$:

*   $R_0 = \{0, 9\}$ (remainder $0$)
*   $R_1 = \{1\}$ (remainder $1$)
*   $R_2 = \{2, 5\}$ (remainder $2$)

We need to select three digits $(d_2, d_3, d_4)$ such that their remainders sum to $1 \pmod 3$. The possible combinations of remainders are $(0, 0, 1)$, $(1, 1, 2)$, and $(2, 2, 0)$.

### Case Analysis

**Case 1: Remainders $(0, 0, 1)$**

There are $3$ possible arrangements for these remainders (e.g., $001, 010, 100$). For each arrangement, the number of ways to choose the digits is:

$$|R_0| 	imes |R_0| 	imes |R_1| = 2 	imes 2 	imes 1 = 4$$

The total for Case 1 is $3 	imes 4 = 12$.

**Case 2: Remainders $(1, 1, 2)$**

There are $3$ possible arrangements for these remainders. The number of ways to choose the digits is:

$$|R_1| 	imes |R_1| 	imes |R_2| = 1 	imes 1 	imes 2 = 2$$

The total for Case 2 is $3 	imes 2 = 6$.

**Case 3: Remainders $(2, 2, 0)$**

There are $3$ possible arrangements for these remainders. The number of ways to choose the digits is:

$$|R_2| 	imes |R_2| 	imes |R_0| = 2 	imes 2 	imes 2 = 8$$

The total for Case 3 is $3 	imes 8 = 24$.

### Final Calculation

By summing the results from our three cases, we obtain the total count of valid integers:

$$12 + 6 + 24 = 42$$

There are exactly **42** such numbers that satisfy the given conditions.

Animated Solution for Mathematics - Permutations and Combinations: The number of numbers greater than 5000, less than 9000 and divisible by 3, that can be formed using the digits 0, 1, 2, 5, 9, if the repetition of the digits is allowed, is ......... .

Enter Numerical Value:

Visualized Solution

Problem Setup & Constraints

Fixing the First Digit

Applying Divisibility Rule

Simplifying the Condition

Categorizing Digits

Valid Remainder Combinations

Case 1: Combination $(0, 0, 1)$

Case 2: Combination $(1, 1, 2)$

Case 3: Combination $(2, 2, 0)$

Final Calculation

The Sigma Insight: Fundamental Principle of Counting

Analyzing the Setup

The Master Equation

The Categorization Strategy

Case Analysis

Final Calculation

Similar Questions

The number of numbers, strictly between 5000 and 10000 can be formed using the digits 1, 3, 5, 7, 9 without repetition, is

The number of five-digit numbers, greater than 40000 and divisible by 5, which can be formed using the digits 0, 1, 3, 5, 7 and 9 without repetition, is equal to

The number of natural numbers less than 7,000 which can be formed by using the digits 0,1,3,7,9 (repetition of digits allowed) is equal to :

The total number of numbers, lying between 100 and 1000 that can be formed with the digits 1, 2, 3, 4, 5, if the repetition of digits is not allowed and numbers are divisible by either 3 or 5 is

Total numbers of 3-digit numbers that are divisible by 6 and can be formed by using the digits 1, 2, 3, 4, 5 with repetition, is ________

Numbers are to be formed between 1000 and 3000, which are divisible by 4, using the digits 1, 2, 3, 4, 5 and 6 without repetition of digits. Then the total number of such numbers is ______.

The total number of three-digit numbers, divisible by 3, which can be formed using the digits 1, 3, 5, 8, if repetition of digits is allowed, is

The number of 5 digit numbers which are divisible by 4, with digits from the set ${1, 2, 3, 4, 5}$ and the repetition of digits is allowed, is ________.

If the digits are not allowed to repeat in any number formed by using the digits 0, 2, 4, 6, 8, then the number of all numbers greater than 10,000 is equal to

Number of 4-digit numbers (the repetition of digits is allowed) which are made using the digits 1, 2, 3 and 5, and are divisible by 15, is equal to ____.

Animated Solution for Mathematics - Permutations and Combinations: The number of numbers greater than 5000, less than 9000 and divisible by 3, that can be formed using the digits 0, 1, 2, 5, 9, if the repetition of the digits is allowed, is ......... .

Enter Numerical Value:

Visualized Solution

Problem Setup & Constraints

Fixing the First Digit

Applying Divisibility Rule

Simplifying the Condition

Categorizing Digits

Valid Remainder Combinations

Case 1: Combination .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (0,0,1)

Case 2: Combination .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (1,1,2)

Case 3: Combination .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (2,2,0)

Final Calculation

The Sigma Insight: Fundamental Principle of Counting

Analyzing the Setup

The Master Equation

The Categorization Strategy

Case Analysis

Final Calculation

Similar Questions

The number of numbers, strictly between 5000 and 10000 can be formed using the digits 1, 3, 5, 7, 9 without repetition, is

The number of five-digit numbers, greater than 40000 and divisible by 5, which can be formed using the digits 0, 1, 3, 5, 7 and 9 without repetition, is equal to

The number of natural numbers less than 7,000 which can be formed by using the digits 0,1,3,7,9 (repetition of digits allowed) is equal to :

The total number of numbers, lying between 100 and 1000 that can be formed with the digits 1, 2, 3, 4, 5, if the repetition of digits is not allowed and numbers are divisible by either 3 or 5 is

Total numbers of 3-digit numbers that are divisible by 6 and can be formed by using the digits 1, 2, 3, 4, 5 with repetition, is ________

Numbers are to be formed between 1000 and 3000, which are divisible by 4, using the digits 1, 2, 3, 4, 5 and 6 without repetition of digits. Then the total number of such numbers is ______.

The total number of three-digit numbers, divisible by 3, which can be formed using the digits 1, 3, 5, 8, if repetition of digits is allowed, is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The number of 5 digit numbers which are divisible by 4, with digits from the set {1,2,3,4,5} and the repetition of digits is allowed, is ________.

If the digits are not allowed to repeat in any number formed by using the digits 0, 2, 4, 6, 8, then the number of all numbers greater than 10,000 is equal to

Number of 4-digit numbers (the repetition of digits is allowed) which are made using the digits 1, 2, 3 and 5, and are divisible by 15, is equal to ____.

Case 1: Combination $(0, 0, 1)$

Case 2: Combination $(1, 1, 2)$

Case 3: Combination $(2, 2, 0)$

The number of 5 digit numbers which are divisible by 4, with digits from the set ${1, 2, 3, 4, 5}$ and the repetition of digits is allowed, is ________.