Mad Factory

Let the volume of the sphere with radius $r$ be equal to $a$ . Let the volume of the sphere with radius $2r$ be equal to $8a$ .

Now, we first realize that in order for the spheres of radius $r$ to have 14 cubic cm of iron left over, $a > 14$ , since no more spheres could have been made from the remaining iron.

Suppose that $x$ spheres of volume $a$ and $y$ spheres of volume $8a$ were made. We can say the following:

$ax + 14 = 8ay + 36$

Rearranging and factoring, we get:

$a(x - 8y) = 22$

Since the volumes of the spheres are all integers, and $a > 14$ , the only possible way we can satisfy the above equation is by letting $a = 22$ and $x - 8y = 1$ .

Therefore, the volume of one of the larger spheres, $8a$ , is equal to $22 * 8$ or $\boxed{176}$ cubic cm.

Let the volume of each block be $V$ . Let the volume of each small sphere be $w$ . Then, each large sphere will have a volume of $8w$ , by the laws of similarity. Let the number of small spheres produced be $x$ and the number of large spheres produced be $y$ .

$\dfrac{V}{w} = x$ with a remainder of $14$ :

$V = wx + 14$ , where $w>14$ since the divisor is always greater than the remainder.

$\dfrac{V}{8w} = y$ with a remainder of $36$ :

$V = 8wy + 36$ , where $8w>36$ since the divisor is always greater than the remainder.

From the two above inequalities, we know that $w>14$ .

Rearranging the two equations:

$V-wx = 14$

$V - 8wy = 36$

Subtracting the first equation from the second,

$wx - 8wy = 22$

$w(x-8y) = 22$

Since, $w$ is given to be an integer, and $x$ and $y$ are by definition integers, $w$ and $x-8y$ must be factors of 22.

The choices we have are:

1. $w = 1, x-8y = 22$

2. $w=2, x - 8y = 11$

3. $w=11, x-8y = 2$

4. $w = 22, x-8y = 1$

But we know that $w>14$ . Hence, $w =22 \Longrightarrow 8w = 8 \times 22 = \boxed{176}$

Firstly, a sphere of radius $2r$ has $2^{3}=8$ times the volume of the sphere with radius $r$ .
Let the volume of each block be $x$ . Then $x-14=rm$ for some positive integer $m$ and $r>14$ . The inequality is because the remainder, $14$ , must be less than the divisor, $r$ . Similarly, $x-36=8rn$ for some positive integer $n$ and $8r>36$ which must hold true if the $r=14$ . Subtracting the second equation from the first, we get $(x-14)-(x-36)=rm-8rn$ Simplifying the left side and factorising the right side gives $22=r(m-8n)$ . This means $r$ divides 22 as $m-8n$ is an integer. So $r$ can only be 1, 2, 11 or 22. However, since $r>14$ , $r=22$ and thus the volume of the larger sphere is $8 \times 22=176$ .

Let, volume of one block is $x$ cubic cm, number of spheres of radii $r$ is $n_1$ and number of spheres of radii $2r$ is $n_2$ .

Form the given information we get, $x=14+n_1\times \frac{4}{3}\pi r^3=36+n_2\times \frac{4}{3}\pi (2r)^3$ .

Solving the equations we get, $r^3=\frac{21}{4\times (n_1-8n_2)}$ .

As $\frac{4}{3}\pi r^3$ is integer, we find $n_1-8n_2 | 22$ . So possible choices of $n_1-8n_2$ are $1,2,11,22$ and for these values possible choices of the volumes for small spheres are $22, 11,2,1$ respectively and volumes for large spheres are $176, 88,16,8$ respectively. From these choices $22$ (for small sphere) and $176$ (for large sphere) is feasible because for small spheres we are left with $14$ cubic cm metal and for large spheres we are left with $36$ cubic cm metal ( $176>36$ and $22>14$ ).

So answer is $\boxed{176}$

volume of small sphere: v and volume of big sphere: 8v
n.v+14 = m.8v + 36 where m and n are integers
v(n-8m) = 22 since v>14 v = 22 and 8v = 176

A be volume of sphere with radius r therefore 8A is the volume of sphere with 2r let x and y be no: of spheres with rad r and 2r respectively xA+14=8yA+36 (eq1) A=22/(x-8y) since all are integers check for x and y for y=1 x=9 that implies A=22

Satisfies eq 1.

Call x the volume of each of the smaller spheres, of each larger sphere is then 8x.
With m,n,x as integers satisfying $x>14$
we have $m*x+14=n*(8x)+36$ $x*(m-8n)=22=1*22=2*11$
$x>14$ so $x=22$
Answer: 8x = $\fbox{176}$

Note: all of the solutions here are wrong; the volume of iron increases as it heats up.