asked Aug 15, 2017 in Other Math Topics by McCquabena Bannor Level 4 User (6,440 points)

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1 Answer

There is a choice of 3 digits for the last digit: 4, 6 or 8.

For the remaining two digits all the given digits are eligible.

Let's start with the hundreds digit. There is a choice of 6 digits,and for the tens digit there are 5 to choose from. Therefore there are 6*5=30 choices for hundreds and tens.

Within the 30 combinations, only 6 are made up entirely of odd numbers; 35, 37, 53, 57, 73, 75. For the accompanying ones digit there are 3 choices, so we have 18 3-digit numbers here.

Within the remaining 24 combinations we have 6 made up entirely of even numbers: 46, 48, 64, 68, 84, 86. That means there's only one even digit for the ones. Therefore in this set we have a further 6 3-digit numbers.

The remaining 18 combinations have a choice of 2 digits for the ones, which gives us 36 3-digit numbers.

The total number of 3-digit numbers is therefore 18+6+36=60.



answered Aug 15, 2017 by Rod Top Rated User (550,780 points)

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