To find the maximum value of the digit 'p' in the eight-digit number $59044p22$ such that it is divisible by 6, we need to apply the divisibility rules for 6.

Understanding Divisibility by 6

A number is divisible by 6 if and only if it meets two conditions:

• It must be divisible by 2.
• It must be divisible by 3.

Checking Divisibility by 2

The rule for divisibility by 2 states that the last digit of the number must be an even digit (0, 2, 4, 6, or 8).

In the number $59044p22$, the last digit is 2.

Since 2 is an even digit, the number $59044p22$ is always divisible by 2, regardless of the value of 'p'.

Checking Divisibility by 3

The rule for divisibility by 3 states that the sum of the digits of the number must be divisible by 3.

Let's calculate the sum of the digits of $59044p22$:

Sum $= 5 + 9 + 0 + 4 + 4 + p + 2 + 2$

Sum $= (5 + 9 + 0 + 4 + 4 + 2 + 2) + p$

Sum $= 26 + p$

For the number to be divisible by 3, the sum ($26 + p$) must be a multiple of 3.

Finding the Maximum Value of 'p'

The variable 'p' represents a single digit, so its possible values are $0, 1, 2, 3, 4, 5, 6, 7, 8, 9$.

We need to find the largest value of 'p' from this set that makes $26 + p$ divisible by 3.

Let's test the possible values of 'p', starting from the largest (9) downwards:

• If $p = 9$, Sum $= 26 + 9 = 35$. ($35 \div 3$ has a remainder of 2)
• If $p = 8$, Sum $= 26 + 8 = 34$. ($34 \div 3$ has a remainder of 1)
• If $p = 7$, Sum $= 26 + 7 = 33$. ($33 \div 3 = 11$). 33 is divisible by 3.

Since we found a value of 'p' (which is 7) that satisfies the divisibility by 3 condition, and we are looking for the maximum value, $p=7$ is the answer.

Other values like $p=4$ (Sum=30) and $p=1$ (Sum=27) also make the number divisible by 3, but 7 is the maximum among these possibilities.

Conclusion

The maximum value of the digit 'p' for the number $59044p22$ to be divisible by 6 is 7.