Find the number of positive three-digit even integers whose digits are among 9, 8, 7, 5, 3, and 1.

The number of positive three-digit even integers whose digits are among 9, 8, 7, 5, 3, and 1 are:

                               36

We are asked to find the number of positive three-digit even integers whose digits are among 9, 8, 7, 5, 3, and 1.

We know that a number is even if the last digit of the number is divisible by 2 i.e. even.

Hence, the only digits among the given digits which is even is: 8

Now, at the first place any of the 6 digits could come up.

( Since, the digits could be repeated)

Also, at the second palace any of the 6 digits could come up.

Hence, the total number of such numbers possible are:

                 6×6×1=36

99

step-by-step explanation:

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