Finding the Maximum Digit 'p' Using Divisibility Rule of 9

The question asks us to find the maximum possible value for the digit 'p' in the eight-digit number 31425p99, such that the number is perfectly divisible by 9.

Understanding the Divisibility Rule for 9

A key rule in number theory states that a number is divisible by 9 if and only if the sum of its digits is divisible by 9. We will use this rule to solve the problem.

Calculating the Sum of Digits

The digits in the given number are 3, 1, 4, 2, 5, p, 9, and 9. Let's find the sum of these digits:

Sum = $3 + 1 + 4 + 2 + 5 + p + 9 + 9$

First, sum the known digits:

$3 + 1 = 4$

$4 + 4 = 8$

$8 + 2 = 10$

$10 + 5 = 15$

$15 + 9 = 24$

$24 + 9 = 33$

So, the total sum of the digits is $33 + p$.

Applying the Divisibility Rule

For the number 31425p99 to be divisible by 9, the sum of its digits, $33 + p$, must be a multiple of 9.

Since 'p' represents a single digit, its value can only be an integer from 0 to 9 (i.e., $0 \le p \le 9$).

Let's determine the possible range for the sum $33 + p$:

• Minimum sum (when $p=0$): $33 + 0 = 33$
• Maximum sum (when $p=9$): $33 + 9 = 42$

We need to find a multiple of 9 that falls within the range [33, 42]. The multiples of 9 are ..., 18, 27, 36, 45, ...

The only multiple of 9 between 33 and 42 is 36.

Determining the Value of 'p'

Therefore, we must have:

$33 + p = 36$

To find 'p', we subtract 33 from both sides:

$p = 36 - 33$

$p = 3$

Conclusion on Maximum Value

The calculation shows that the only possible value for the digit 'p' that makes the number 31425p99 divisible by 9 is 3. Since there is only one possible value, this value is also the maximum value.

Thus, the maximum value of p is 3.