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Riddle
By Yatir Halevi on Saturday, January 12, 2002 - 07:41 pm:

Take a 3-digit number, where the hundredth digit is the biggest (eg. 321), and subtract from it the same number only backwards (123).
Add the digits of the answer (198) and you will always get 18.
I have a proof for it.
But I would like to see how many different proofs people will find from as many different areas of mathematics.

Good Hunting,
Yatir

By Anthony Cardell Tony on Saturday, January 12, 2002 - 08:40 pm:

This is probably the same proof as you have:

Let the three digit number be represented by 100a+10b+c where a,b,c are integers from 0 to 9. Now, if we do the subtraction we are just doing (100a+10b+c)-(100c+10b+a)=99(a-c). Since the hundredth digit is the largest, 99(a-c) is positive. Also, since 99 is divisible by 9, the sum of the digits of 99(a-c) must also be divisible by nine (well known divisibility rule). Let the sum of the digits of 99(a-c) be 9x. Now, 99 is also divisible by eleven, and so if we let 99(a-c)=100d+10e+f, then d+f-e=0 (because of the divisibility rule for eleven), so d+f=e. Since d+f+e=9x, then 2e=9x. It is easy to see from this that we must have e=9 and x=2 since e can only be from 0-9. Therefore, the sum of the digits is 9x=9(2)=18.

By Yatir Halevi on Saturday, January 12, 2002 - 09:19 pm:

My proof is almost the same.
I to get 99(a-c). But I said that the middle digit has to be 9 (very easy to prove, and I've done this in a different method than yours) So inorder for 99(a-c) to be divisible by 9 (that it obviously does). a+c has to be equal to 9.
So we get 9+9=18

Yatir

By Brad Rodgers on Saturday, January 12, 2002 - 10:05 pm:

I have a slightly different proof. We know that upon subtracting, we have

(a-c-1)102+9×101+(10+c-a)

so the coeffecients are (a-c-1), 9, and 10+c-a; when we add these, we obviously get 18.

Brad

By Yatir Halevi on Sunday, January 13, 2002 - 04:38 pm:

Brad, Yours is the other proof I had.

Yatir