The total number of three-digit numbers, divisible by 3, which can be formed using the digits 1, 3,

Q: Wait, why do we reject the negative root here? Or rather, why can't we just use simple probability?

That is a great question! In combinatorics, especially for JEE Advanced, we avoid probability because we need the exact count of integers. We aren't looking for a ratio; we are looking for the total number of valid outcomes. We use case-based counting because it ensures we don't double-count or miss any valid combinations. Stay systematic, and the answer will reveal itself!

Q: How do I know I have found all the combinations for the $AAB$ case?

The best way to be sure is to use the modular arithmetic we discussed. For the $AAB$ case, the sum is $2A + B \equiv 0 \pmod{3}$. By testing each possible value for $A$ (which can be $1, 3, 5,$ or $8$), you can solve for the required remainder of $B$. This leaves no room for guessing and ensures you catch every valid pair.

Q: Why do we divide by $2!$ in the $AAB$ case?

Excellent catch! When we have a set like $\{5, 5, 8\}$, the number of ways to arrange these digits is not just $3!$ because the two $5$s are identical. If we swapped the two $5$s, the number would remain the same. Therefore, we divide by the factorial of the number of identical items ($2!$) to remove those redundant arrangements. It is a fundamental rule of permutations with repetition!

Question

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

Options:

(A) 21

(B) 20

(C) 22

(D) 18

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We are tasked with finding the total number of $3$-digit numbers divisible by $3$ using the digits $\{1, 3, 5, 8\}$, with repetition allowed. Let the three digits be $D_1, D_2,$ and $D_3$.

The fundamental condition for divisibility by $3$ is that the sum of the digits must be a multiple of $3$. Mathematically, we require:

$$D_1 + D_2 + D_3 = 3k, \quad 	ext{where } k \in \mathbb{Z}$$

To solve this systematically, we categorize the numbers based on the repetition of digits: all identical, exactly two identical, or all distinct.

### Case 1: The Uniform Numbers ($AAA$)

In this scenario, all three digits are the same. We test each digit from the set $\{1, 3, 5, 8\}$ to see if $3A$ is a multiple of $3$:

*   If $A=1$, sum $= 3$ (Valid)
*   If $A=3$, sum $= 9$ (Valid)
*   If $A=5$, sum $= 15$ (Valid)
*   If $A=8$, sum $= 24$ (Valid)

Since all four digits satisfy the condition, we have $4$ valid numbers: $111, 333, 555,$ and $888$.

### Case 2: The Mixed Numbers ($AAB$)

Here, exactly two digits are identical. We require the sum $2A + B$ to be a multiple of $3$. We test the combinations:

*   If $A=1$, $2(1)+B = 2+B$. For this to be a multiple of $3$, $B$ must be $1$ (already covered in Case 1) or $4$ (not in set).
*   If $A=3$, $2(3)+B = 6+B$. For this to be a multiple of $3$, $B$ must be $3$ (already covered in Case 1).
*   If $A=5$, $2(5)+B = 10+B$. For this to be a multiple of $3$, $B$ must be $8$ (since $18$ is a multiple of $3$).
*   If $A=8$, $2(8)+B = 16+B$. For this to be a multiple of $3$, $B$ must be $5$ (since $21$ is a multiple of $3$).

We have two distinct sets: $\{5, 5, 8\}$ and $\{8, 8, 5\}$. For each set, the number of arrangements is given by:

$$\frac{3!}{2!} = 3$$

Thus, we have $2 	imes 3 = 6$ ways.

### Case 3: The Distinct Numbers ($ABC$)

Finally, we consider the case where all three digits are different. We look for triplets $\{A, B, C\}$ from $\{1, 3, 5, 8\}$ such that their sum is a multiple of $3$:

*   $\{1, 3, 5\}$: Sum $= 9$ (Valid)
*   $\{1, 3, 8\}$: Sum $= 12$ (Valid)
*   $\{1, 5, 8\}$: Sum $= 14$ (Invalid)
*   $\{3, 5, 8\}$: Sum $= 16$ (Invalid)

Since all digits in these sets are distinct, each set can be arranged in $3! = 6$ ways. With two valid sets, we have:

$$2 	imes 6 = 12 	ext{ ways}$$

### The Final Tally

By summing the results from our three logical cases, we arrive at the total count:

$$	ext{Total} = 4 (	ext{Case 1}) + 6 (	ext{Case 2}) + 12 (	ext{Case 3}) = 22$$

The total number of $3$-digit numbers divisible by $3$ using the given digits is **$22$**.

Animated Solution for Mathematics - Permutations and Combinations: 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

Select Answer:

Visualized Solution

Understanding the Objective

The Divisibility Rule for $3$

Case 1: All Digits Identical ( $AAA$ )

Case 2: Two Identical Digits ( $AA B$ )

Case 2: Arrangements

Case 3: All Digits Distinct ( $A B C$ )

Case 3: Arrangements

Final Summation

Conclusion

The Sigma Insight: Fundamental Principle of Counting

Analyzing the Setup

Case 1

Case 2

Case 3

The Final Tally

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