Primary School Logic Problem: Finding Possibilities for Balloon Spots

[benachie]

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?

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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!

 

[Dickfore]

If there are x balloons with 2 spots, and y balloons with 5 spots, then the total number of spots is N, and is related to the unknown quantities by:
$$2x + 5y = N$$
This is a linear equation with 2 unknowns. Therefore your problem is underconstrained and should not be expected to have a unique solution. In fact, if x and y were real numbers, then the possible solutions all lie on a straight line in the xy-plane.

Now, your unknowns are integers (and not smaller than 2). This problem is then known as a Diophantine equation.

There is an algorithm for finding all the solutions of such problems. It has been implemented on this website.