Here's one for Keltin

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Wow, that’s pretty cool. I’ve seen other math type puzzles like this where I can usually guess the formula use, but this one is a real stumper!! Thanks for sharing that! :cool:
 
I'm going to play it again - with the exact same numbers I used the first time -

let me try again - my calculator is so small I probably hit a wrong number.....

be right back.............:cool:
 
ok - it was my fingers on the calculator - and every time that dog barks at the end my dog goes nuts!!!!!!!!!!!!!!!!!! :LOL:
 
I tried to trick it by using 1234 and 1324. After subtracting, my number was 90, so I could only circle 9 leaving me just a 0 to type in......and it got it; it picked 9! Hmmm.....there’s a mathematical reason for this..........
 
My dear wife has worked it out. The numbers in your final number always add up to 9 (or multiple) so the digits you give the puzzle are subtracted from 9, the remainder is the number you circled.
 
Heh, I figured it out. :cool:


Spoiler Alert!
Don't read any further unless you want to know how it is done!








This is a trick based on a magician's technique called a “Force” and that’s when the magician causes ("forces") you to select a specific item or group of items, yet you think you had total free will.

The trick tells you to pick a 3 or 4 digit number. Then it has you mix up those digits for another number. Then you subtract the two, and the you then pick a number from the result of that subtraction. That’s the “force”. You think you are picking a random number, but it’s not. There is a finite set of numbers that will happen because of this limited subtraction scheme.

Let’s look at a 3 digit combo of the numbers 1, 2, & 3. The largest number you can make is 321, and the smallest is 123. The total unique numbers that you can make are 6 which are:

The three hundreds = 321, 312
The two hundreds = 231, 213
The one hundreds = 132, 123

Total unique subtractions with all number combinations are 15. This means there are only 15 new numbers that you can eventually choose (circle) your number from that he must “guess”. And of those 15, two combinations are not allowed because the result is a single 9. Further, you aren’t allowed to choose a 0 which would have stumped this trick, but by disallowing the zero, it’s easy to make a look-up table to cross reference the remaining digits you type in to the limited numbers that can actually be made.

Here are all the possible subtractions that can happen starting with the largest number of 321 and working down:

321
312 = 9 <--not allowed, single digit!
213 = 108 <--can’t choose a Zero
231 = 90 <--can’t choose a Zero
132 = 189
123 = 198

312
213 = 99
231 = 81
123 = 189
132 = 180 <--can’t choose a Zero

231
213 = 18
123 = 108 <--can’t choose a Zero
132 = 99

213
123 = 90 <--can’t choose a Zero
132 = 81

132
123 = 9 <--Not allowed, single digit!


So, if you did 312 – 123 = 198 and you circle the 9, you have to type in 1 & 8. By typing in 1 & 8, that means you had a three digit remainder. Look at that list, of all the 3 digit remainders with a 1 & 8 in them (remember 0 is not allowed), the only number left for you to have chosen is a 9.

Or, say you chose 312 – 132 = 180. You can’t choose the 0, so you choose 1. Now you have to type in 8 & 0. Of all the 3 digit numbers with 8 & 0, the third number is always....you guessed it........1.

The interesting thing is, those results are the same no matter what 3 digits you picked! That’s because you are working with the same 3 digits for the subtraction, thus the magnitude of difference between the two numbers is always the same, no matter what three you pick. 987 – 789 = 198 which is the same as 321 -123 = 198! The magnitude of the difference is always the same, no matter the 3 digits you pick to start with, so the resulting remainder is always the same.

The same thing applies to a 4 digit number, with the only difference being there are more remainders, but it is still a finite list of numbers!

Mystery solved!
 
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