math problem?

Question

please show work if possible

1. the numbers, 999, 777, and 111 contain a total of nine digits. cross out one or more of the digits and the sum of the numbers left need to equal eighteen. how would the answer to this puzzle change if the directions change to: cross out one or more of the digits and the sum of the digits left need to equal eighteen?

2. how many three-digit positive integers have three different even digits?

Answer 1

PROBLEM 1:

If the numbers have to add up to eighteen, there is one way to end up with 18:
7 + 11

If the digits have to add up to eighteen, there are two ways to end up with 18:
9 + 9
9 + 7 + 1 + 1

That's it.

PROBLEM 2:

Let's start with the even digits {0, 2, 4, 6, 8}

For the first digit we have *four* choices (assuming you can't have a leading zero in a three digit integer. {2, 4, 6, 8}

After that we have four choices for the next digit and 3 choices for the last digit.

4 x 4 x 3 = 48 ways

Here they are:

204, 206, 208, 240, 246, 248, 260, 264, 268, 280, 284, 286,
402, 406, 408, 420, 426, 428, 460, 462, 468, 480, 482, 486,
602, 604, 608, 620, 624, 628, 640, 642, 648, 680, 682, 684,
802, 804, 806, 820, 824, 826, 840, 842, 846, 860, 862, 864

Answer 2

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