# Primary school problem!

• Nov 23rd 2012, 12:38 AM
benachie
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?

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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!
• Nov 23rd 2012, 01:04 AM
MarkFL
Re: Primary school problem!
This is how I would go about it:

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

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

Next, I would let:

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

$\displaystyle 3(5n+a)+5(b-3n)=k$

$\displaystyle 3a+5b=k$

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

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

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

$\displaystyle 3(7)+5(2)=31$ hence:

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

$\displaystyle x=5n+7$ and $\displaystyle y=2-3n$

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

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

But, $\displaystyle n=-2$ 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 $\displaystyle k$.