What Is the Mechanism Behind the Fido Puzzle?

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The Fido Puzzle involves selecting a 3 or 4 digit number, rearranging its digits, and subtracting the smaller from the larger to yield a result that retains a specific property related to "casting out 9s." This method shows that the sum of the digits of both the original and rearranged numbers will yield the same result when reduced, which is crucial for solving the puzzle. The subtraction always results in a positive integer, and the outcome's digits can be analyzed to deduce a missing digit based on the casting out 9s principle. This principle ensures that the final digit can be determined by knowing the sum of the remaining digits. The puzzle's design cleverly utilizes modular arithmetic to maintain consistency in the results.
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Try to 4build equations. 4567 would be 4*1000 + 5*100 + 6*10 +7 etc...

t*1000+u*100+v*10+w - w*1000 + v*100 + u*10 + t ... 1st equation
 
I'm in 11th grade. I don't get it.
 
May I please remind you linux kid that the tutorials sections is not the place to post questions.
 
Looks to me like a variation on "casting out 9s". That is, take a number and add its digits. If the sum is larger than 9, do the same to the sums.

For those who don't want to go to that website: You pick a 3 or 4 digit number (you are encouraged to chose one with several different digits). Say 3214, for example. Write another number made from those same digits rearranged: 4132, say. Now subtract the smaller of those two numbers from the larger: 4132- 3214= 918.

You are now instructed to choose a NON-ZERO digit from that number, say 1, and type in the remaining digits, 98. The site then tells you the digit you picked, in this case 1.

If you "cast out 9s" with the number 3214 you get 3+2+1+4= 10. Since that is larger than 9, do it again: 1+ 0= 1. If you do the same thing with the rearranged number, 4132, since you have exactly the same digits to add, you get the same thing: 1. Now the crucial point is that the result of casting out 9s is "preserved" by the usual arithmetic operations: adding, subtracting (as long as the subtraction leaves a positive integer), multiplying, dividing (as long as the division leaves a positive integer). To prove that you need to write the number as a sum of powers of 10 as NWScience suggested.

Since you are instructed to "subtract the smaller number from the larger" you will always get a positive integer and since the original number and it rearranged form have the same "casting out 9s" number, the number you get when you subtract MUST give 9 (same as 0 since this is essentialy "modulo 9") after casting out 9s! In this case, the number we got by subtracting was 918:9+ 1+ 8= 18= 9. Knowing all except 1 digit of this number, since we know their "casting out 9s", we can find that last digit. Here, if we picked 1 and told the site that 98 remained, it is easy to calculate that 9+ 8= 17, 1+ 7= 8 so the remaining digit must be 9- 8= 1. If we had instead picked the digit 9 we would have told the site that we were left with 18: 1+ 8= 9. This is the "ambiguous case" since in "casting out 9s" 0 and 9 are equivalent: that;s why we were specifically instructed to choose a NON-ZERO digit!
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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