Animated Solution for Mathematics - Permutations and Combinations: 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 ____.

A Combinatorial Journey

Welcome, future engineer! Today, we are going to unravel a beautiful problem that sits at the intersection of number theory and combinatorics. It is not just about finding an answer; it is about understanding the constraints that define the structure of numbers.

Imagine you are tasked with building a 4-digit number using only the digits $\{1,2,3,5\}$, with the strict requirement that the resulting number must be divisible by 15. Let us break this down step-by-step.

The Divisibility Constraint

First, let us look at the number 15. It is a composite number, the product of two coprime factors: 3 and 5.

This means that for any number to be divisible by 15, it must simultaneously satisfy the divisibility rules for both 3 and 5. If we ignore either, we lose the game.

For divisibility by 5, the rule is elegant and simple: the last digit must be either 0 or 5. We look at our toolkit—the set $\{1,2,3,5\}$.

Since 0 is not available, the last digit of our 4-digit number is locked. It must be 5. We have already fixed one position!

The Sum Condition

Now, we turn our attention to the divisibility rule for 3. A number is divisible by 3 if and only if the sum of its digits is a multiple of 3.

Let our 4-digit number be represented by the digits $d_1,d_2,d_3,d_4$. We know $d_4=5$. Therefore, the sum of the digits is $d_1+d_2+d_3+5$.

For this to be divisible by 3, we must have:

where $k$ is some integer. Let us rearrange this to isolate the sum of the first three digits:

We can rewrite $3k-5$ as $3(k-2)+1$. Let $m=k-2$. Thus, the condition becomes:

In plain English, the sum of the first three digits must leave a remainder of 1 when divided by 3. This is the key that unlocks the entire problem.

The Combinatorial Hunt

Now, we must systematically find all combinations of three digits from our set $\{1,2,3,5\}$ that satisfy this sum condition. The minimum possible sum is $1+1+1=3$, and the maximum is $5+5+5=15$.

The possible sums that leave a remainder of 1 when divided by 3 are 4, 7, 10, and 13. Let us list the combinations for each sum:

Sum = 4: The only combination is $\{1,1,2\}$. Sum = 7: We have $\{1,1,5\}$, $\{1,3,3\}$, and $\{2,2,3\}$. Sum = 10: The only combination is $\{2,3,5\}$. Sum = 13: The only combination is $\{3,5,5\}$.

The Final Tally

Now, we calculate the number of arrangements for each combination. For combinations with two identical digits, the number of arrangements is given by:

For the combination with distinct digits, it is $3!=6$.

1. For $\{1,1,2\}$ (Sum 4): 3 ways. 2. For $\{1,1,5\}$ (Sum 7): 3 ways. 3. For $\{1,3,3\}$ (Sum 7): 3 ways. 4. For $\{2,2,3\}$ (Sum 7): 3 ways. 5. For $\{2,3,5\}$ (Sum 10): 6 ways. 6. For $\{3,5,5\}$ (Sum 13): 3 ways.

Adding these up: $3+3+3+3+6+3=21$.

There we have it! By carefully applying the rules of divisibility and combinatorics, we have found that there are exactly 21 such numbers. Keep this systematic approach in your arsenal, and no JEE problem will ever be able to stand against you!