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Cross out a number - any number - and I'll tell you what it


I hand you a sheet of paper with the following instructions:

[1] Write down a 6-digit number. Say you write 352687

[2] Scramble the digits. Say you come up with 762853

[3] Subtract the smaller from the larger. You get: 762853 - 352687 = 410166

[4] Cross out one of the digits [but not a zero, cuz it's basically not there anyway.] Say you cross out a 6.

[5] Scramble the remaining digits. Say you get: 61401

[6] Tell me the digits. You say 6 1 4 0 1.

What are the odds that I will say, "you crossed out a 6"?

Odds are 100%.

Anybody can say "You crossed out a 6"

If you got that answer, congratulations, you're a nit-picker, and nobody likes to talk to you. :P

==> OK then, the REAL question is,

==> What are the odds that I will correctly tell you the number that you crossed out? B))

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Clue: How do the numbers 6 1 4 0 1 give away that 6 was crossed off?

Another case: 753487 - 345877 = 407610

this time cross out a 7.

scramble the remaining digits and tell me 6 0 0 1 4.

How do those numbers give away that a 7 was crossed off?

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I don't have any mathematical proof, bt it appears fro the examples shown that all 6 digits of the result add up yo 18, so it is simple math 18- 12 in the first post and 18-11 in the second post.

Now Bananova can explain how / why we always come up with a digit total of 18, or is this just coincidence?

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Now Bananova can explain how / why we always come up with a digit total of 18, or is this just coincidence?
You got it. Nice going!

And you resisted the red herring of 0 1 4 and 6 appearing in both cases.

The trick works because of two interesting properties of the number 9.

[1] scrambling a number's digits changes its value by a multiple of 9.

[2] adding the digits of a multiple of 9 gives another multiple of 9.

So the solution is to subtract the sum of the given digits from the next

higher multiple of 9. It just happened to be 18 both times here.

And the prohibition of not crossing out a zero is to distinguish

it from the case of crossing out a 9.

Now, can you prove these two interesting properties?

[1] Start with the simplest case: move a 1 from the units place to the tens place.

You've added 9 to the value: 10 - 01 = 9.

Move it from the units place to the hundreds place.

you've added 99 to the value: 100 - 001 = 99.

And so on for moving to any other place in the number = always a multiple of 9.

Move a number other than 1, say 7.

You change the above results by a factor 7, but it's still a multiple of 9:

70 - 07 - 63 = 9*7

700 - 007 - 693 = 99*7

Pair switching produces sums of multiples of 9.

Scrambling all digits is a series of pair switches.

[2] Work the above backwards:

Start with 9.

Decrease it by 1 and put the 1 in some other place: 18, 108, 1008, etc.

Two things happen. The digits still add to 9;

and, by the above reasoning, the new number differs from 9 by a multiple of 9.

You can create any desired multiple of 9 by successive changes of digit values in this manner.

Thus any multiple of 9 adds digit-wise to a multiple of 9.

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There are entire books- entirely scholarships- devoted to the number 9. It's my favorite number and i could probably load you up with so many "9-magic-tricks" and "9-riddles" it would make your brains explode! lol

but i'm content to sit back and watch other ppl solve em :D

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