## PROBLEM 33. SNOWBALLS

A boy made two snowballs, one of which has a diameter twice as large as the other. He brought them into a warm room and let them melt. Since only the surface of the snowball is exposed to the warm air, assume that the melting rate is proportional to the surface area. When half of the volume of the larger snowball has melted, how much of the smaller snowball will remain?
Solution. We will prove that the assumption that the melting rate of a snowball is proportional to the area of its surface leads to a surprising result, namely that the radius decreases at a constant rate regardless of the size of the snowball. Therefore, both radii will decrease by the same amount.

The volume and surface area are respectively $V=4 \pi r^{3} / 3$ and $S=4 \pi r^{2}$.

Denoting time by $t$, we get that the melting rate is

$$
\frac{d V}{d t}=\frac{4}{3} \pi \cdot 3 r^{2} \cdot \frac{d r}{d t}=4 \pi r^{2} \frac{d r}{d t}
$$

Since this rate is proportional to the surface area $S=4 \pi r^{2}$, we have

$$
4 \pi r^{2} \frac{d r}{d t}=k\left(4 \pi r^{2}\right)
$$

where $k$ is a constant. Hence,

$$
\frac{d r}{d t}=k
$$

as claimed.

58

Suppose initially the radii were $2 r$ and $r$. Then the volume of the larger snowball is

$$
V=\frac{4}{3} \pi(2 r)^{3}=\frac{32}{3} \pi r^{3}
$$

After half of it has melted, it will have a volume of

$$
V=\frac{16}{3} \pi r^{3}=\frac{4}{3} \pi(r \sqrt[3]{4})^{3}
$$

from which its radius is $r \sqrt[3]{4}$. Each radius will thus decrease by $(2-\sqrt[3]{4}) r$. Then the smaller snowball will have a radius of

$$
r-(2-\sqrt[3]{4}) r=r(\sqrt[3]{4}-1)
$$

and its volume will be

$$
V=\frac{4}{3} \pi r^{3}(\sqrt[3]{4}-1)^{3}
$$

which is very close to $1 / 5$ of the original volume $\left((\sqrt[3]{4}-1)^{3} \approx 0.2027\right)$.

## PROBLEM Z4. DIGITS OF NUMBERS FROM ONE to a BILLION

What is the sum of all the digits used in writing all the numbers, the first of which is one and the last is a billion?

Solution. Adding zero, we can form half a billion pairs of numbers:

(0; 999999999), (1; 999999998), (2; 999999997), ...

..., (499999998; 500000001), (499999999; 500000000).

The sum of the digits in each pair will be $9 \cdot 9=81$.

If we add 1 to the sum of the digits for the unaccounted number 1000000000, we get the sum quite simply:

$$
500000000 \cdot 81+1=40500000001 .
$$