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Math Help - Primary school problem!

  1. #1
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    Primary school problem!

    My son who is at primary school was given the following homework:

    "Finding all possibilities Logic Problem"


    Jessica and Simon were blowing up balloons for Garreths birthday. There were at least two of each. Some balloons had three spots and some had 5 spots.

    There were 31 spots altogether.

    Q: How many balloons had three spots and how many had 5 spots?

    What if there were 24 spots?
    What if there were 65 spots?

    ---------------------------------

    OK, so knowing that there were two of each to start with (5+5+3+3=16 spots) we only need to establish the possible permutations for the remaining 15 ie 3 x 5 spot balloons or 5 x 3 spot balloons

    For 24 spots there is only one possible answer (24 - 16 = 8 spots = 1 x 5 spot balloon and 1 x 3 spot balloon

    For 65 there are a few permutations 65-16=49

    8 x 5 + 3 x 3 = 49
    2 x 5 + 13 x 3 = 49
    5 x 5 + 8 x 3 = 49

    But can we be sure that we have found all the possible answers? Is there a formulae for testing and am I posting this question in the right place!
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  2. #2
    MHF Contributor MarkFL's Avatar
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    Re: Primary school problem!

    This is how I would go about it:

    Let 2\le x be the number of balloons with 3 spots and 2\le y be the number of balloons with 5 spots. We then have:

    3x+5y=k where k in the total number of spots.

    Next, I would let:

    x=5n+a and y=b-3n and so there results:

    3(5n+a)+5(b-3n)=k

    3a+5b=k

    Now, we want to find a specific instance that works for a particular k.

    In the case of k=31, I would begin taking multiples of 5 away from 31 until I have a multiple of 3. In this case:

    31-2(5)=21=3(7) and so we find:

    3(7)+5(2)=31 hence:

    a=7,\,b=2 and thus:

    x=5n+7 and y=2-3n

    We know n=0 produces a valid solution (x,y)=(7,2), but n=1 does not, so going in the other direction we find:

    n=-1\rightarrow(x,y)=(2,5)

    But, n=-2 produces an invalid solution, so we know we have found the only two solutions that work.

    A similar process can be done for the other two values of k.
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