Total numbers of 3-digit numbers that are divisible by 6 and can be formed by using the digits 1, 2,

Q: Why do we start with divisibility by 2? Can't we just start with 3?

Excellent question! In combinatorics, we always prioritize the most restrictive constraint. Divisibility by 2 only cares about the last digit, which immediately narrows our search space. Divisibility by 3 involves all digits, making it a 'global' constraint. Always solve the 'local' constraint first to simplify the 'global' one.

Q: Wait, do I need to worry about the order of $a$ and $b$?

Yes, absolutely! In a 3-digit number, the position matters. The number $132$ is distinct from $312$. When you list your pairs $(a, b)$, treat them as ordered pairs. If $a+b=4$, the pair $(1, 3)$ is distinct from $(3, 1)$.

Q: Is there a formula for this, or do I have to list them?

While there are advanced generating function methods, for a small set like this, systematic listing is the most robust and error-proof method. JEE Advanced rewards clarity and systematic thinking over complex formulas that might lead to calculation errors.

Question

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 ________

Sigma Percentile · Accepted Answer

### Analyzing the Setup

To form a 3-digit code using the set $\{1, 2, 3, 4, 5\}$ with repetition allowed, we must satisfy the condition of divisibility by 6. A number is divisible by 6 if and only if it is divisible by both 2 and 3.

The divisibility rule for 2 is a local constraint: the last digit $c$ must be even. Given our set $\{1, 2, 3, 4, 5\}$, the possible values for $c$ are $2$ or $4$.

The divisibility rule for 3 is a global constraint: the sum of the digits $S = a + b + c$ must be a multiple of 3. We define this as:

$$a + b + c = 3k, \quad k \in \mathbb{Z}$$

### The First Universe ($c = 2$)

When the unit digit is $c = 2$, the condition becomes $a + b + 2 = 3k$. Since $a, b \in \{1, 2, 3, 4, 5\}$, the sum $a + b$ ranges from $2$ to $10$. Consequently, $a + b + 2$ ranges from $4$ to $12$.

The multiples of 3 in this range are 6, 9, and 12. We analyze the scenarios for $a + b$:

1. $a + b + 2 = 6 \implies a + b = 4$: The pairs $(a, b)$ are $(1, 3), (3, 1), (2, 2)$. (**3 cases**)
2. $a + b + 2 = 9 \implies a + b = 7$: The pairs $(a, b)$ are $(2, 5), (5, 2), (3, 4), (4, 3)$. (**4 cases**)
3. $a + b + 2 = 12 \implies a + b = 10$: The pair $(a, b)$ is $(5, 5)$. (**1 case**)

Summing these, we find $3 + 4 + 1 = 8$ valid numbers for this universe.

### The Second Universe ($c = 4$)

When the unit digit is $c = 4$, the condition becomes $a + b + 4 = 3k$. The sum $a + b + 4$ ranges from $6$ to $14$.

The multiples of 3 in this range are 6, 9, and 12. We analyze the scenarios for $a + b$:

1. $a + b + 4 = 6 \implies a + b = 2$: The pair $(a, b)$ is $(1, 1)$. (**1 case**)
2. $a + b + 4 = 9 \implies a + b = 5$: The pairs $(a, b)$ are $(1, 4), (4, 1), (2, 3), (3, 2)$. (**4 cases**)
3. $a + b + 4 = 12 \implies a + b = 8$: The pairs $(a, b)$ are $(3, 5), (5, 3), (4, 4)$. (**3 cases**)

Summing these, we find $1 + 4 + 3 = 8$ valid numbers for this universe.

### Final Calculation

Since the two universes are mutually exclusive, we apply the rule of sum to find the total count.

The total number of valid 3-digit codes is:

$$8 + 8 = 16$$

**The total number of valid codes is 16.**

Animated Solution for Mathematics - Permutations and Combinations: 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 ________

Enter Numerical Value:

Visualized Solution

Understanding the Constraints

Divisibility by $6$

Constraint 1: Divisibility by $2$

Constraint 2: Divisibility by $3$

Case 1: Unit Digit $c = 2$

Counting Pairs for $c = 2$

Case 2: Unit Digit $c = 4$

Counting Pairs for $c = 4$

Final Calculation

The Sigma Insight: Fundamental Principle of Counting

Analyzing the Setup

The First Universe ( $c = 2$ )

The Second Universe ( $c = 4$ )

Final Calculation

Similar Questions

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 total number of four digit numbers such that each of the first three digits is divisible by the last digit, 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

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 ......... .

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 ____.

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 ________.

The number of 3-digit odd numbers, whose sum of digits is a multiple of 7, is ______.

The number of natural numbers lying between 1012 and 23421 that can be formed using the digits 2, 3, 4, 5, 6 (repetition of digits is not allowed) and divisible by 55 is ______.

The number of integers greater than 6,000 that can be formed, using the digits 3, 5, 6, 7 and 8, without repetition, is :

Animated Solution for Mathematics - Permutations and Combinations: 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 ________

Enter Numerical Value:

Visualized Solution

Understanding the Constraints

Divisibility by .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 6

Constraint 1: Divisibility by .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 2

Constraint 2: Divisibility by .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 3

Case 1: Unit Digit .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } c=2

Counting Pairs for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } c=2

Case 2: Unit Digit .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } c=4

Counting Pairs for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } c=4

Final Calculation

The Sigma Insight: Fundamental Principle of Counting

Analyzing the Setup

The First Universe ( .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } c=2)

The Second Universe ( .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } c=4)

Final Calculation

Similar Questions

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 total number of four digit numbers such that each of the first three digits is divisible by the last digit, 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

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 ......... .

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 ____.

.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 ________.

The number of 3-digit odd numbers, whose sum of digits is a multiple of 7, is ______.

The number of natural numbers lying between 1012 and 23421 that can be formed using the digits 2, 3, 4, 5, 6 (repetition of digits is not allowed) and divisible by 55 is ______.

The number of integers greater than 6,000 that can be formed, using the digits 3, 5, 6, 7 and 8, without repetition, is :

Divisibility by $6$

Constraint 1: Divisibility by $2$

Constraint 2: Divisibility by $3$

Case 1: Unit Digit $c = 2$

Counting Pairs for $c = 2$

Case 2: Unit Digit $c = 4$

Counting Pairs for $c = 4$

The First Universe ( $c = 2$ )

The Second Universe ( $c = 4$ )

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 ________.