Understanding the Divisibility Rule for 4
To determine if a number is divisible by 4, we only need to look at the number formed by its last two digits. If the number formed by the last two digits is divisible by 4, then the entire number is divisible by 4.
Applying the Rule to 87893P4
In the given 7-digit number, $87893P4$, the last two digits form the number $P4$. Here, 'P' represents a single digit from 0 to 9.
According to the divisibility rule for 4, the number $87893P4$ will be divisible by 4 if and only if the number $P4$ is divisible by 4.
Finding the Largest Digit for P
We need to find the largest possible digit that can replace P such that the two-digit number $P4$ is divisible by 4. Let's test the possible digits for P, starting from the largest (9) downwards:
- If P = 9: The number formed is 94. Is 94 divisible by 4? $94 \div 4 = 23.5$. No, 94 is not divisible by 4.
- If P = 8: The number formed is 84. Is 84 divisible by 4? $84 \div 4 = 21$. Yes, 84 is divisible by 4.
Since we are looking for the largest number that should replace P, and we found that P=8 works, this is our answer. We don't need to check smaller digits like 2 or 0, although they would also make the number divisible by 4 (forming 24 and 04 respectively).
Conclusion
The largest digit that can replace P in the number $87893P4$ to make it divisible by 4 is 8.
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Important Questions from Divisibility and Remainder
If the 8-digit number 888x53y4 is divisible by 72, then what is the value of (7x + 2y), for the maximum value of y?
If all positive divisors of 132 are arranged in descending order, then what digit will be at unit place of first divisor ?
If 3 2019 is divided by 10, then what is the remainder?
The number 3798125P369 is divisible by 7. What is the value of the digit P?
Consider all 3-digit numbers (without repetition of digits) obtained using three non-zero digits which are multiples of 3. Let S be their sum.
Which of the following is/are correct?
1. S is always divisible by 74.
2. S is always divisible by 9.
select the correct answer using the code given below:
