Sam

1.) There is an unusual dartboard. Each dart scores either 3 points or 8 points. Suppose you can throw as many darts as you like, and your sore is obtained by adding all the 3s and 8s together. Make a list of all the scores that are impossible to attain.


2.) Arrange the digits 1 through 6 like the model below so that the multiplication works out correctly.

ab * cde = fgh

The first correct answers will get 10 points.

Jack

All of the impossible scores?

1,2,4,5,7,10,13, and 18?

And of course all of the negative integers.

Aaron

1) Ignoring zero and negative numbers, the easiest way to do this is to prove that all scores above a certain number are possible. Let's try 15.

15 is obviously possible. 3 * 5.
16 is possible: 2 * 8.
17 is possible, too: 8 + 3 * 3.

But if those three are possible, then any other number can be obtained by adding multiples of 3 to one of them! Between the sequences (15, 18, 21, 24...), (16, 19, 22, 25...), and (17, 20, 23, 26...), all the positive integers above 15 are covered!

So we just have to worry about numbers under 15. To check each number:

Is it a multiple of 3 or 8? If so, it's possible.
Else, when you subtract 8, is it a multiple of 3? If so, it's possible.
Else, when you subtract 16, is it a multiple of 3? If so, it's possible.
Otherwise, it's impossible.

Quickly applying this gives us our list of impossibles:

1, 2, 4, 5, 7, 10, 13.

2) Er, there are eight letters. A B C D E F G H. How do we arrange six digits that way?

Also, even 10 * 100 gives a four-digit answer, 1000. There's no way to make two digits times three digits give a three-digit result, as far as I can see.

Izzy

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The Saint

These aren't riddles, they seem more like homework,

Homework this way~~~~~~~~>lol...!