form the smallest 5 digit number using 9 8 7 6 and 5 so that it is divisible by 5
Answers
Answer:
The smallest 5 digit number that is divisible by 5 = 67895
Concept
The range of five-digit numbers is ten thousand to ninety-nine thousand nine hundred and ninety-nine. At the ones, tens, hundreds, thousands, and ten thousand places in these numbers are integers. These locations accept any positive integer, with the exception of 0 at the tens of thousands place. This is because adding a 0 here will make the number a 4-digit one; for example, 04356 will become 4356.
Given
9 8 7 6 and 5 are the numbers given.
The smallest digit number should be divisible by 5.
Find
We are asked to frame the smallest 5 digit number using the given numbers which should be divisible by 5.
Solution
A number must end with the digits 0 or 5 in order to be divisible by 5. Since there is only one value between 0 and 5, the number must end in 5.
The smallest number must come first in order for the number to be produced, however 5 is already utilized as the final digit. So, it is assumed that 6 is the first digit. The smallest number is made up of the last three digits, 7, 8, and 9, which are ordered in ascending order.
Two numbers, 67895 and 68795, for instance, are smaller in the first case because the thousands place has a smaller integer.
Hence, the smallest 5 digit number using 9,8,7,6,5 which is divisible by 5 is 67895
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