Anonymous
Anonymous asked in Science & MathematicsMathematics · 9 years ago

Math Number Puzzle Combination/Riddle?

You can ONLY use the following numbers:

11.3,

11.6,

12.9,

12.7,

11.5,

9.8,

9.4,

8.5,

8.4

The Challenge: The goal is to come up with as many SMALL combinations of numbers (using the numbers above), when averaged, equals to 11.5

For example: 11.3 + 11.6 + 11.6 / 3 = 11.5 (average)

The is a fun challenge. I've come up with eight ways. Can you beat me? I'll post results later.

Update:

You can not use multiples of the value that you average.

So if I used 11.3 + 11.6 + 11.6 = 34.5 , of course you could just multiply this value by 2 ( = 69 ) and divide by 6 to obtain the same average, 11.5

As a result, you cannot use multiples of the value that you add up to, to get your average.

Another example:

((11.3 x 6) + 12.7) / 7 = 11.5 I would end here. I would not take the average, multiply it by 2, and divide by 14 to obtain "another way of getting 11.5". Can't use multiples. Hope that helps.

3 Answers

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  • 9 years ago
    Favorite Answer

    In a computer search, I found that there are exactly 186 solutions using at most 9 different numbers from your list (possibly multiple times each), though I didn't get rid of multiples (there were only a dozen or so by my inspection).

    I found myself again wanting easy access to higher forms of iteration (eg. iterate over subsets with or without order, over multisets; over cartesian products; I'm sure there's a huge variety of useful higher order iteration). I would include my complete list if it would fit. Instead, I picked a couple dozen pseudo-randomly.

    (7*11.3 + 1*12.9 + 1*11.5) / 9

    (7*11.3 + 1*12.9) / 8

    (6*11.3 + 1*12.7 + 2*11.5) / 9

    (4*11.3 + 1*12.9 + 2*12.7 + 1*11.5 + 1*8.5) / 9

    (4*11.3 + 1*12.9 + 2*12.7 + 1*8.5) / 8

    (3*11.3 + 6*11.6) / 9

    (3*11.3 + 3*11.6 + 2*12.7 + 1*9.4) / 9

    (3*11.3 + 1*11.6 + 1*12.9 + 1*12.7 + 2*11.5 + 1*9.4) / 9

    (3*11.3 + 1*11.6 + 1*12.9 + 1*12.7 + 1*11.5 + 1*9.4) / 8

    (3*11.3 + 1*11.6 + 1*12.9 + 1*12.7 + 1*9.4) / 7

    (3*11.3 + 1*11.6 + 3*12.7 + 1*11.5 + 1*8.4) / 9

    (3*11.3 + 1*11.6 + 3*12.7 + 1*8.4) / 8

    (2*11.3 + 1*12.9 + 2*12.7 + 1*11.5 + 2*9.8) / 8

    (2*11.3 + 1*12.9 + 2*12.7 + 2*9.8) / 7

    (1*11.3 + 5*11.6 + 2*12.9 + 1*8.4) / 9

    (1*11.3 + 1*12.9 + 3*12.7 + 1*9.8 + 1*8.4) / 7

    (3*11.6 + 2*12.9 + 1*11.5 + 1*8.4) / 7

    (3*11.6 + 2*12.9 + 1*8.4) / 6

    (3*11.6 + 1*12.9 + 4*11.5 + 1*9.8) / 9

    (3*11.6 + 1*12.9 + 3*11.5 + 1*9.8) / 8

    (3*11.6 + 1*12.9 + 2*11.5 + 1*9.8) / 7

    (3*11.6 + 1*12.9 + 1*11.5 + 1*9.8) / 6

    (3*11.6 + 1*12.9 + 1*9.8) / 5

    (3*11.6 + 4*12.7 + 1*9.4 + 1*8.5) / 9

    (2*11.6 + 2*12.9 + 1*12.7 + 2*11.5 + 2*9.4) / 9

    (2*11.6 + 2*12.9 + 1*12.7 + 1*11.5 + 2*9.4) / 8

    (2*11.6 + 2*12.9 + 1*12.7 + 2*9.4) / 7

    (1*11.6 + 1*12.9 + 4*12.7 + 3*9.4) / 9

    (1*11.6 + 5*12.7 + 1*8.5 + 1*8.4) / 8

    (2*12.9 + 3*12.7 + 2*9.8 + 1*8.5) / 8

    (4*12.7 + 2*11.5 + 1*9.8 + 1*8.4) / 8

    (4*12.7 + 1*11.5 + 1*9.8 + 1*8.4) / 7

    (4*12.7 + 1*9.8 + 1*8.4) / 6

    (1*11.5) / 1

  • 9 years ago

    Numbers Difference from mean

    11.3, -0.2

    11.6, 0.1

    12.9, 1.4

    12.7, 1.2

    11.5, 0.0

    9.8, -1.7

    9.4, -2.1

    8.5, -3.0

    8.4 -3.1

    I have thought over the problem. As you do not allow multiples you should allow the use of any reading more than once. otherwise as Josh swanson says there is no limit numberfof ways they could be easily infinite!

  • Anonymous
    9 years ago

    I've found an infinite number of ways.

    -After additional details-

    I've still got an infinite number of ways.

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