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We have got our yearly festival tomorrow. So my two ancient aunts are making these totally cool sweets. I was there to help them and just check our what I found.

My 1st aunt is really great at cooking while the 2nd is still trying to catch it. Well the amount of sweets they made just proves it. Both my aunts cooked sweets as much as the square of their children. And the total amount of sweets that both of them made is as the cube of the no. of children my 2nd aunt has.

Given that the whole family, about 30 people would be coming to the get together tomorrow.

So, How many kids do each of my aunts have?

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We can set up an equation for a problem like this:

$x$ is the number of children the first aunt has and $y$ is the number of children the second aunt has. Then the first aunt made $x^2$ sweets and the second $y^2.$ Their total is $x^2+y^2.$

Since the total is the cube of $y,$ we can say $x^2+y^2 = y^3.$

Then do some simplification:

Subtract $y^2$ from both sides to get $x^2 = y^3-y^2.$ Now factor out $y^2$ from the right to get $x^2 = y^2(y-1).$

So what does this tell us?

Well, the left side is a perfect square, so the right side must also be a perfect square. Since $y^2$ is already a perfect square, we must have $y-1$ also be a perfect square. Given the limitation of having $30$ people, this tells us that $y-1$ equals either $1,$ $4,$ $9,$ $16,$ or $25.$ So $y$ equals one of $2,5,10,17,26.$

But...

We also must have $x+y \le 30.$ With a little testing, we can see that $y=2$ works as $y^2(y-1) = 2^2 = 4$ so $x=2$ also. $y=5$ also works, as $y^2(y-1) = 25\cdot 4 = 100$ which means $x=10.$ But $y=10$ won't work since $y^2(y-1) = 100\cdot9 = 900.$ This means that $x=30$ but then we'd have more than 30 people at the reunion.

So either....

They both have $2$ kids or aunt 1 has $10$ kids and aunt 2 has $5$ kids. To check both:

$2^2+2^2=8$ which is equal to $2^3$ and $10^2 +5^2 = 125$ which is equal to $5^3.$ Given that the first aunt is better at cooking, I'm going to say that the first aunt has $10$ kids and the second $5.$

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Your first aunt has 10 children and your second has 5.

Explanation:

Your description leads to the equation $x^2+y^2=y^3$ (where $x$ is the number of children of your first aunt and $y$ is the number of children of your second aunt.

Looking for integer solutions to this, we can see pairs such as $(0,0)$, $(0,1)$, $(2,2)$, $(10,5)$, and $(30,10)$. However, we've been told to expect approximately 30 people. $(10,5)$ means 15 children, plus a number of other family members (a minimum of 18 people including you and the aunts). $(30, 10)$ is a minimum of 43 people, so the former answer is the closest.

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They can have

both 2 children

Because

both cooked 2^2=4 sweets, making it 8 in total which is the number of children either aunt has, cubed.

But then,

how does it prove that the 1st aunt it better than the second at cooking I don't know O_o

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  • $\begingroup$ Taste of sweets, I guess. :P $\endgroup$ – Sid Oct 2 '17 at 12:58
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I think the right answer is:

The 1st Aunt cooked 100 sweets because she has 10 children.

The second aunt

Cooked 25 sweets because she has 5 children.

Finally,

Total no. Of sweets=125 which is the cube of the number of children of the second aunt.

This also satisfies the other condition:

1st Aunt is obviously a better cook because she cooks 4 times as much as the second Aunt. And if we include the children and everything, the total number of people might come out to be 30.

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They have

10 and 5

children, so they made

100 and 25 which is 125 (=5^3) sweets.

If we think everybody mentioned here will come, than the next minimum will be too high:

at 30 and 10 children, 30(aunt 1's children) + 10( aunt 2's children) + 2(aunts) + 1(story teller) = 43 minimum person.

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