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How many 4-digits positive integers (that is, integers between 1000 and 9999, inclusive) have only even digits divisible by 5?A) 80 B) 100 C) 125 D) 200D) 500

Last updated date: 17th Sep 2024
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Hint: In this question we are going to find out how many 4-digits positive integers having only even digits that are 0,2,4,6 and 8. So we have to arrange them in a 4-digit such that it should be divisible by 5, for that we will use the concept of Factorials to calculate the number of ways of taking the 5 even number in the 4 places of a 4-digit number. And the formula used is given below. The formula of using the factorials is just multiplying the number of ways from a given set of objects.

First, we divide the 4 digit number into 4 places that are from the right ones, tens, hundreds, and thousands.
So in this problem, we have to make a four-digit number from even numbers which are divisible by 5. So we have to start from the right-hand side to find the number of even numbers that should be placed in the position to get the desired answer.
Total number of objects is {0, 2, 4, 6, and 8}
On ones place only 0 should be placed as there is only that digit, which makes the number divisible by 5,
So the total number of ways to fill the one’s place is 1
Now moving to the tens place,
Tens place can be any number from the given sets of even number so all the number can be taken in the tens place,
So the total number of ways to fill the one place is 5
Now moving to the hundreds place,
Hundreds place be any number from the given sets of the number so all number can be taken in the hundreds place,
So the total number of ways to fill the hundreds place is 5
Now moving to the thousands place,
Thousands of places can be filled by 4 numbers from a given set of even numbers that are {2, 4, 6, 8} here 0 is not taken as it makes the number into a 3-digit number like 0265 which is a 3-digit number.
So the total number of ways to fill the thousands place is 4
Now we will the concept of factorial to find the number of ways to make a 4-digit from even numbers which are divisible by 5,
We will multiple all the ways we calculated to fill the respective places
So the total number of ways will be =$4 \times 5 \times 5 \times 1$
Now multiply the above numbers we get,
=100

$\therefore$ The total number of ways to make a digit number from the even number which should be divisible by 5 is 100. Hence, Option (B) is correct.

Note:
Use the correct way to solve the problem,
The one’s digit, for all the numbers that have to be divisible by $5$, must be a $0$ or a $5$. Since the problem states that we can only use even digits, the last digit must be $0$. From there, there are no other restrictions since the divisibility rule for $5$ states that the last digit must be a $0$ or a $5$. So there are $4$ even digit options for the first number then $5$ for the middle $2$.
So when we have to do $4 \times 5 \times 5 \times 1 = 100$