1. ## Problem 17

1)Take any three digit number with different digits. Reverse the order of the digits. And find the difference between these two numbers. Consider this result as a three digit number, if you get three digits good, if not pretend there are zeros in front. Reverse the order of this number and add to this number.
Highlight below to see what you have,
1089.
Explain why it works.

2)Prove that for any $\displaystyle n,m>1$ the inequality:
$\displaystyle [(n+m-1)!]^2\leq (2n-1)!(2m-1)!$
Is always satisfied.
And only equality when $\displaystyle n=m$.

2. Originally Posted by ThePerfectHacker
1)Take any three digit number with different digits. Reverse the order of the digits. And find the difference between these two numbers. Consider this result as a three digit number, if you get three digits good, if not pretend there are zeros in front. Reverse the order of this number and add to this number.
Highlight below to see what you have,
1089.
Explain why it works.
Just to make sure I have the right procedure:
893
893 - 398 = 495
495 + 594 = 1089
Which is the same sum as every other possibility?

-Dan

3. Originally Posted by topsquark
Just to make sure I have the right procedure:
893
893 - 398 = 495
495 + 594 = 1089
Which is the same sum as every other possibility?

-Dan
Hello, Dan,

not quite. You allways get 1089. I only don't know why.

EB

4. Originally Posted by topsquark
Just to make sure I have the right procedure:
893
893 - 398 = 495
495 + 594 = 1089
Which is the same sum as every other possibility?

-Dan
You got the right procedure!

5. Originally Posted by ThePerfectHacker
1)Take any three digit number with different digits. Reverse the order of the digits. And find the difference between these two numbers. Consider this result as a three digit number, if you get three digits good, if not pretend there are zeros in front. Reverse the order of this number and add to this number.
Highlight below to see what you have,
1089.
Explain why it works.
It isn't pretty, but it works. And I don't know why the digits have to be different (unless all he means is that we can have 441, but not 444.) The only restriction I can really find is that the first digit and the last digit aren't the same.

Given any 3 digit number abc = 100*a + 10*b + c, where a and c are not equal.

Then abc - cba = 99|a - c|.

Since this is a single digit number times 99 we know that 99|a - c| is a 2 or three digit number.

If we have a 2 digit number, then the number is 99. 099 + 990 = 1089.

If we have a 3 digit number the middle digit is a 9 and since 99|a - c| is divisible by 9 we know that the sum of the first and last digits is 9. (The sum of the digits of a number divisible by 9 is a multiple of 9. The only possibility for the sum of the digits of a 3 digit number divisible by 9 is 18.) Thus adding 99|a - c| and the number with its digits in reverse order is of the form:
d9e + e9d
where e + d = 9.
Thus this number has a last digit of 9.
The 10s place is 9 + 9 = 18, so we have 89 as the final digits, carry the 1.
d + e = 9 and we carry a 1, so 10.
Thus the sum is 1089.

-Dan

6. The first problem I will not answer just yet because I am in college and do not want do upload pictures. I will update this post when I return back home if you want to see the solution.

2)This is my own inequality which I am really proud of. The reason is because I think the derivation is elegant. This seems like an elementary problem, and perhaps it is, I just did not find an elementary derivation. And yet is uses two seemingly unrelated things, to it, and to each other.

Definition: The gamma function is defined $\displaystyle s>0$ as, $\displaystyle \Gamma(s)=\int_0^{\infty}e^{-t}t^{s-1}dt$.

In the Calculus tutorial there is some material on the Gamma function if you want to read it.
The fact we need is that $\displaystyle \Gamma(n+1)=n!$.

For this derivation we need to introduce another similar function, let us call it the "Ron" function.

Definition: The Ron function is defined $\displaystyle s>0$ as, $\displaystyle \rho(s)=\int_0^{\infty} e^{-2t}t^{s-1}dt$.

Theorem: The Ron function satisfies, $\displaystyle \rho(s+1)=\frac{1}{2}s\rho(s)$
Proof: Integration by parts on $\displaystyle u'=e^{-2t}$ and $\displaystyle v=t^{s-1}$. And the same stuff as with the Gamma function.

Repeated application of the Ron function we have $\displaystyle \boxed{ \rho(n+1)=\frac{n!}{2^{n+1}} }$

That is the first seemingly unrelated thing to this problem. The second one is the Cauchy-Swartz Inequality on general inner product spaces.

Basically an inner product space (from linear algebra) is a real vector space for which an "inner product" was defined. That is a binary operation among vectors into reals, written as $\displaystyle <\bold{u},\bold{v}>$. To be an inner product space it needs to satisfy,
$\displaystyle <\bold{u},\bold{v}>=<\bold{v},\bold{u}>$
$\displaystyle <\bold{u}+\bold{w},\bold{v}>=<\bold{u},\bold{v}>+< \bold{w},\bold{v}>$.
$\displaystyle <\bold{u},\bold{u}>\geq 0$ and only zero when $\displaystyle \bold{u}=\bold{0}$.

Definition: Norm is defined as $\displaystyle ||\bold{u}||=\sqrt{<\bold{u},\bold{u}>}$

Here is my favorite inequality,
Cauchy-Swartz Inequality: For an inner product space $\displaystyle |<\bold{u},\bold{v}>|\leq ||\bold{u}||\cdot ||\bold{v}||$. And only equality when both are equal.

Now a vector space we will consider are all continous functions on $\displaystyle [0,\infty)$ over the reals. The inner product will be defined as,
$\displaystyle \int_0^{\infty} f(x)g(x) dx$.
We see this satisfies the Inner product space axioms. Hence we can rely upon the Cauchy-Swartz Inequality (after squaring both sides).
$\displaystyle \left(\int_0^{\infty} f(x) g(x) dx\right)^2\leq \int_0^{\infty} f^2(x) dx \int_0^{\infty}g^2(x) dx$

We know that, for $\displaystyle n,m>1$ integers we have,
$\displaystyle n!=\int_0^{\infty}e^{-t} t^n dt$ and $\displaystyle m!=\int_0^{\infty}e^{-t} t^m dt$
Thus, we will consider the functions $\displaystyle f(t)=e^{-t}t^n$ and $\displaystyle g(t)=e^{-t}t^m$.

By the inequality we have,
$\displaystyle \left( \int_0^{\infty} e^{-t}t^n e^{-t}t^m dt \right)^2 \leq \int_0^{\infty}e^{-2t}t^{2n}dt\cdot \int_0^{\infty} e^{-2t}t^{2m} dt$
$\displaystyle \left( \int_0^{\infty} e^{-2t}t^{n+m} dt \right)^2 \leq \int_0^{\infty} e^{-2t}t^{2n} dt\cdot \int_0^{\infty} e^{-2t}t^{2m} dt$
Using the Ron function we have,
$\displaystyle \rho^2 (n+m)\leq \rho (2n)\rho(2m)$
Using the Ron identity for integers,
$\displaystyle \frac{[(n+m-1)!]^2}{2^{n+m}}\leq \frac{(2n-1)!(2m-1)!}{2^n \cdot 2^m}$
Clearing denominators,
$\displaystyle [(n+m-1)!]^2\leq (2n-1)!(2m-1)!$
(And only equality for $\displaystyle n=m$ gaurentted by the Cauchy-Swartz Inequality).

Note, if you want you can make it a little more elegant you can write,
$\displaystyle \Gamma (n+m)\leq \Gamma(n)\Gamma(m)$
And this holds for non-integer values as well.
Thus for example set $\displaystyle n=m=.5$.
Then we have,
$\displaystyle 1\leq \frac{\sqrt{\pi}}{2}\cdot \frac{\sqrt{\pi}}{2}$
$\displaystyle \pi < 4$
And easy way to show that pi does not exceede 4.

--------------
2) This is a classic number thing. I seen it a few times in stupid magic trick books. My first solution was based on topsquarks and Soroban's but then I found that there is a much simpler way to show it, it almost makes me laught how simple it can be done.
You are given a number ABC without lose of generality assume CBA < ABC thus, you subtract CBA from ABC. And furthermore you will need to borrow in order to succesfully do substraction. Show below.

I would like to mention that I once did this with ABCD and it is supprising that it can change (I forgot the exact conditions that I got) but it is either 10890 or 10898 (I think). And that the probability is equally likely.

7. Hello, ThePerfectHacker!

1) Take any three digit number with different digits.
Reverse the order of the digits, and find the difference between these two numbers.
Consider this result as a three-digit number.
If you get three digits good; if not, pretend there is a zero in front.
Reverse the order of this number and add to this number.
Highlight below to see what you have,
*1089*
Explain why it works.

Let the number be: .$\displaystyle N \:=\:100H + 10T + U$. .WNLOG, assume $\displaystyle H > U.$

To reverse and subtract, some "borrowing" is necessary.

We have: .$\displaystyle \begin{array}{cc}100H +10T + U\\ 100U + 10T + H\end{array}$

which becomes: .$\displaystyle \begin{array}{cc}100(H - 1) \\ 100U\end{array} \begin{array}{cc} + \\ + \end{array} \begin{array}{cc}10(T + 9) \\ 10T\end{array} \begin{array}{cc} + \\ + \end{array} \begin{array}{cc}(U - H + 10) \\ U \end{array}$

Subtract: .$\displaystyle 100(H - U - 1) + 10(9) + (U - H +10)$
Reverse: .$\displaystyle 100(U - H + 10) + 10(9) + (H - U - 1)$

Add: .$\displaystyle 100(9) + 10(18) + 9 \;=\;\boxed{1089}$