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Two Math Questions for all of you geniuses...

kitkat2013kitkat2013 32 replies22 threads Junior Member
Blue Book Test 3 Part 8: #13
In a certain game, each token has one of three possible values: 1 point, 5 points, or 10 points. How many combinations of these token values are worth a total of 17 points?
Correct answer: Six

How do you solve this other than just listing them out and hoping you don't miss one? Or is that it?

Blue Book Test 3 Part 8: #16
A cube with volume 8 cubic centimeters is inscribed in a sphere so that each vertex of the cube touches the sphere. What is the length of teh diameter, in centimeters, of the sphere?
Correct answer: 2(sqrt3) - about 3.46

I solved this by approximating, but I'd like to know how to truly solve it.

Thank you in advance.
edited February 2012
2 replies
Post edited by kitkat2013 on
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Replies to: Two Math Questions for all of you geniuses...

  • 2400Man2400Man 140 replies34 threads Junior Member
    For the second one, since it tells you that the cube's volume is 8, then that means the height, width, and length of the cube is 2. Since it tells you that each vertex of the cube touches the sphere, then that means that the diameter of the sphere is actually the diagonal of the cube. Think of the problems with a square inscribed in a circle and the diagonal of the square being the diameter of the circle. It's just that type of problem except in 3D! Now you just solve for the diagonal of the cube by using this formula: square root of x^2+y^2+z^2. Thus the square root of 4+4+4=the square root of 12 which can be simplified into 2square root of 3. Tada!
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  • fogcityfogcity 3171 replies57 threads Senior Member
    For the first one there is no simple formula that will help. Enumeration however is very straightforward and if you proceed methodically you won't miss a combination. Start with combinations that must include a 10, then those that must include at least one-five and no-tens, and finally those that include only ones.

    Start by finding out how many combinations have a 10.

    10 + 5 + two-ones
    10 + seven-ones

    Then how many combinations have one or more 5s (we've already dealt with 10s above ... so none of these have a ten)

    5 + twelve-ones
    two-5s + 7-ones
    three-5s + 2-ones

    Then how many combinations have only ones


    Try a similar problem -- same rules -- except the sum is 26 and in addition to 10, 5 and 1 point pieces you also have 25 point pieces.
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