If the probability that a randomly chosen 6-digit number formed by using digits 1 and 8 only is a mu

Q: Wait, why do we split the number into $1000A + B$? Can't we just divide the number by 7?

You could, but that would be a nightmare! With 64 possibilities, long division is a trap. By using $N = 1000A + B$, we leverage the fact that $1000 \equiv -1 \pmod 7$. It turns a massive calculation into a simple subtraction: $B - A$. It is the difference between brute force and elegance.

Q: How did you know $1 \equiv 8 \pmod 7$ was the key?

In competitive exams like JEE Advanced, whenever you see a constraint involving a prime number like 7, always look for patterns in modular arithmetic. The digits 1 and 8 are congruent to 1 mod 7. This is not a coincidence; it is a structural property of the problem designed to reward those who look for symmetry.

Q: Why is $n_1$ restricted to $\{0, 3, 6\}$?

The sum of digits $S$ must be a multiple of 3. We derived $S = 48 - 7n_1$. Since $48$ is a multiple of 3, we need $7n_1$ to be a multiple of 3. Since 7 is coprime to 3, $n_1$ must be a multiple of 3. Since $n_1$ is the count of 1s in a 6-digit number, it must be between 0 and 6. Hence, 0, 3, and 6.

Question

If the probability that a randomly chosen 6-digit number formed by using digits 1 and 8 only is a multiple of 21 is $p$, then $96p$ is equal to ____.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing before a 6-digit number, constructed entirely from the digits 1 and 8. You are asked to determine the probability that this number is a multiple of 21.

We have 6 slots, and each slot has 2 choices (1 or 8). By the fundamental principle of counting, the total number of outcomes is:

$$2^6 = 64$$

This is our sample space. In the realm of JEE Advanced, we do not use brute force; we use insight.

### The Divisibility Trap

The number 21 is a composite number, which is our first clue. To be a multiple of 21, a number must be a multiple of both 3 and 7.

### The Modulo 7 Magic

This is where the problem transforms from a calculation into a revelation. We split the 6-digit number $N$ into two 3-digit blocks, $A$ and $B$, such that $N = 1000A + B$.

Applying modulo 7, since $1000 \equiv -1 \pmod{7}$, we find:

$$N \equiv -A + B \pmod{7}$$

The number is divisible by 7 if and only if $B - A$ is divisible by 7. Note that $1 \equiv 1 \pmod{7}$ and $8 \equiv 1 \pmod{7}$.

In the world of modulo 7, the digits 1 and 8 are identical. This means any 3-digit block $A$ or $B$ formed by these digits will always be congruent to $111 \pmod{7}$.

Since $111 = 7 	imes 15 + 6$, both $A$ and $B$ are congruent to 6. Therefore:

$$B - A \equiv 6 - 6 = 0 \pmod{7}$$

Every single one of our 64 numbers is divisible by 7. The constraint of 7 is satisfied for all cases.

### The Sum of Digits

Now, we only need to satisfy the condition for 3. A number is divisible by 3 if the sum of its digits $S$ is a multiple of 3.

Let $n_1$ be the number of 1s and $n_8$ be the number of 8s. Since there are 6 digits, $n_8 = 6 - n_1$. The sum is:

$$S = 1(n_1) + 8(n_8) = n_1 + 8(6 - n_1) = 48 - 7n_1$$

For $S$ to be divisible by 3, we require $48 - 7n_1 \equiv 0 \pmod{3}$. Since 48 is a multiple of 3, we need $-7n_1 \equiv 0 \pmod{3}$, which simplifies to $n_1 \equiv 0 \pmod{3}$.

Thus, $n_1$ must be 0, 3, or 6.

### The Final Tally

We count the favorable cases:

1) If $n_1 = 0$, we have $\binom{6}{0} = 1$ way.

2) If $n_1 = 6$, we have $\binom{6}{6} = 1$ way.

3) If $n_1 = 3$, we have:

$$\binom{6}{3} = \frac{6 	imes 5 	imes 4}{3 	imes 2 	imes 1} = 20 	ext{ ways}$$

The total favorable outcomes are $1 + 1 + 20 = 22$. The probability $p$ is:

$$p = \frac{22}{64} = \frac{11}{32}$$

Finally, we calculate $96p$:

$$96 	imes \frac{11}{32} = 3 	imes 11 = 33$$

We have arrived at the answer, **33**, not by grinding through numbers, but by understanding the structure of the problem. That is the JEE way.

Animated Solution for Mathematics - Probability: If the probability that a randomly chosen 6-digit number formed by using digits 1 and 8 only is a multiple of 21 is $p$ , then $96 p$ is equal to ____.

Enter Numerical Value:

Visualized Solution

Understanding the Constraints

Calculating Total Possible Numbers

Condition for Divisibility by 3

Sum of Digits

Identifying Possible Counts of Digit 1

Condition for Divisibility by 7

Applying Modulo 7

The Key Insight: $1 \equiv 8 (mod 7)$

Evaluating Blocks A and B

Every Such Number is Divisible by 7

Counting Favorable Cases: $n_{1} = 0$ and $n_{1} = 6$

Counting Favorable Cases: $n_{1} = 3$

Final Probability and $96 p$

The Sigma Insight: Classical Definition of Probability

Analyzing the Setup

The Divisibility Trap

The Modulo 7 Magic

The Sum of Digits

The Final Tally

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