## Task 20/61

For the Klingenberg power plant in Berlin-Rummelsburg, two new chimneys were built. Each of them consists of a concrete casing in the form of a hollow frustum of a cone with the following dimensions:

Lower inner diameter $d_{u}=10.00 \, \text{m}$, upper inner diameter $d_{o}=7.50 \, \text{m}$, lower outer diameter $D_{u}=11.20 \, \text{m}$, upper outer diameter $D_{o}=7.80 \, \text{m}$, height $H=140.00 \, \text{m}$. This casing was lined with glass wool, diatomaceous earth, and brick.

a) How many cubic meters of concrete were required for each of the two chimney casings?

b) What is the weight $G$ of each of the two chimney casings? The specific weight $\gamma$ of the concrete used is assumed to be $\gamma=2.4 \, \text{M p/m}^3$.

c) What pressure does the chimney casing exert on the foundation?

a) The masonry volume is obtained as the difference between the total volume of the structure and the enclosed hollow space. Both are frustums of cones in this case. The formula for the volume of a frustum of a cone is

$$
V=\frac{\pi h}{3}\left(r_{1}^{2}+r_{1} r_{2}+r_{2}^{2}\right)
$$

where $h$ is the height of the frustum, $r_{1}$ and $r_{2}$ are the radii of the base and top circles, respectively.

For the masonry volume, this results in

$$
V=\frac{\pi h}{3}\left(R_{1}^{2}+R_{1} R_{2}+R_{2}^{2}\right)-\frac{\pi h}{3}\left(r_{1}^{2}+r_{1} r_{2}+r_{2}^{2}\right)
$$

with $R_{1}=\frac{D_{u}}{2}=5.60 \, \text{m}, R_{2}=\frac{D_{o}}{2}=3.90 \, \text{m}, r_{1}=\frac{d_{u}}{2}=5.00 \, \text{m}$, and $r_{2}=\frac{d_{o}}{2}=3.75 \, \text{m}$. This yields $V \approx 1555 \, \text{m}^3$.

b) The weight of a body is given by $G=V \cdot \gamma$, resulting in $G \approx 3732 \, \text{M p}$.

c) The pressure $p$ is calculated as the quotient of the effective force (in this case, the weight $G$) and the area over which the force acts. The area $F$ (the ground plan of the chimney casing) is an annulus with an outer radius $R_{a}=5.60 \, \text{m}$ and an inner radius $R_{i}=5.00 \, \text{m}$. It is $F \approx 20 \, \text{m}^2$, so for the pressure $p \approx 18.66 \, \text{kp/cm}^2$. This calculation assumes that the entire load of the chimney casing is evenly distributed over the cross-section.

I wrote 513 and should now set the number "reversed" below it, 315, and subtract it from the first, the difference 198 (which Jörg did not know!) 12 was added to, and the sum 210 was divided by 70; the quotient 3 had to be multiplied by 12, which resulted in 36.

Now Jörg shone with the latest knowledge we had since the last math lesson and demanded that the square root be taken from the product: $\sqrt{36}=6$.

With that, the trick was over, and he announced that we had obtained 6.

Our astonishment was great! He was pressed to say how he did it, and he especially wanted to demonstrate his art again, which he was happy to do.

The matter troubled me all day, and in the afternoon, I sat down and pondered until I discovered the mathematical regularity that enabled Jörg to perform this "math trick" and proved it algebraically.

So how must one proceed? Our task is as follows:

Write any three-digit number whose hundreds place is two greater than the units place. Subtract from it the number obtained by reversing the order of the digits in the original number. Add 12 to the result; then, in sequence, perform the following further arithmetic operations with the respective results: division by 70; multiplication by 12, and take the square root! The result is 6.

1. How is it possible to predict the result of the arithmetic operations for any number as the initial value?
2. Is a general solution possible where the hundreds place is $n$ greater than the units place? $(n=1,2, \ldots, 9)$
1. If $a=100 x+10 y+z$ is any three-digit number, then according to the condition stated in the problem, $x=z+2$, so $a=100(z+2)+10 y+z=100 z+200+10 y+z$.

For the number with the digits reversed, we have: $b=100 z+10 y+x=100 z+10 y+z+2$. Subtracting $b$ from $a$, we get: $a-b=198$. We see that through the subtraction, all the digits of the number $a$ fall out of the calculation, and the result is always 198, regardless of the choice of the number $a$. The further calculations serve only to obscure this fact.

2. Generally, if $x=z+n$ with $n=1,2, \ldots, 9$,

$$
a-b=100 z+100 n+10 y+z-(100 z+10 y+z+n)=100 n-n=99 n
$$

The result $99 n$ of course includes the special case $1(n=2, 99 n=198)$. On this basis, a mathematical "magic trick" can be built without much memorization.

## Solution by Ernst Hennig:

Let the units digit be $x$, the tens digit $y$, and the hundreds digit $x+n$ with $n=1,2,3, \ldots, (9-x)$. The number is thus

$$
100(x+n)+10 y+x
$$

and the number with the digits reversed is $100 x+10 y+(x+n)$. The difference between the two numbers is $D=$ $100 n-n=99 n$.

This difference thus has a constant value, independent of the choice of digits, $99 n$, which depends only on the magnitude $n$.

From here, Jörg can choose two paths. The simpler one is as follows:

1. Jörg adds the number $10+n$ (i.e., $11,12,13, \ldots$) to the difference not named by him. This results in

$$
99 n+10+n=100 n+10=10(10 n+1)
$$

which gives 110, 210, 310, ... He then divides this number by $10 n+1$, i.e., by 11, 21, 31, ... The result is always 10.

2. A bit more challenging for the mathematical magician is the second path, which represents the full generalization of the number game (but only makes sense for $n \geq 2$).
3. Requirement: Add the number 100(n-2)+10(n-1)+n (i.e., $012,123,234, \ldots$) to the difference $D$. Result: $210(n-1)$.
4. Requirement: Divide the result by 70! Result: $3(n-1)$, i.e., $3,6,9, \ldots$
5. Requirement: Multiply the last result by 12(n-1)! The resulting number is $36(n-1)^{2}$.
6. Requirement: Take the square root of the last result! Final result: $6(n-1)$, i.e., $6,12,18, \ldots$