thats intricate

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Its NOT intricate!

I GOT IT.

You take any no. and jumble its digit to get another no. Then subtract smaller no. from bigger no and get a no. say X.

FACT: If you make sum of all digits in X it has to be either 18 or 27. Verify. I'm working on general proof but its hard.

If you tell computer 3 digits, its trivial to guess the 4th one. If sum of 3 digits is less than 18, it will be subtracted from 18 to find 4th digit. If sum of X is more than 18, it will subtracted from 27 to get 4th one. (Its strange that sum of digits in X can't exceed 27, can be verified easily, proof is difficult though)

E.g.
If you got X=2088 (i.e. 3211-1123) and tell compu 8,0,2.... 4th digit is 18-(2+8)=8

Enjoy!
-Vinod
 
vinodks said:
I GOT IT.

You take any no. and jumble its digit to get another no. Then subtract smaller no. from bigger no and get a no. say X.

FACT: If you make sum of all digits in X it has to be either 18 or 27. Verify. I'm working on general proof but its hard.

If you tell computer 3 digits, its trivial to guess the 4th one. If sum of 3 digits is less than 18, it will be subtracted from 18 to find 4th digit. If sum of X is more than 18, it will subtracted from 27 to get 4th one. (Its strange that sum of digits in X can't exceed 27, can be verified easily, proof is difficult though)

E.g.
If you got X=2088 (i.e. 3211-1123) and tell compu 8,0,2.... 4th digit is 18-(2+8)=8

Enjoy!
-Vinod
Click to expand...


no matter what number you select, the final number you get will always be a multiple of 9 and the computer knows it.

rest is simple
 
Ture. For a no of 3 or 4 digits its 18 or 27... Do you have proof for any number of digits?... it can be verified for couple of digits but do you have general analytical proof?

-vinod

arunshamli said:
no matter what number you select, the final number you get will always be a multiple of 9 and the computer knows it.

rest is simple
Click to expand...
 
Last edited:
vinodks said:
Ture. For a no of 3 or 4 digits its 18 or 27... Do you have proof for any number of digits?... it can be varified for couple of digits but do you have general analytical proof?

-vinod
Click to expand...


This is true for any digit number and not only for 3 or 4 digit numbers. in this particular puzzle you can start with any digit number and it will be able to give you the correct answer. aur ye jo 3digits ya 4 digits bolta hai ...woh sirf confuse karne ke liye.....

let me try to explain it


It is a well known fact that if a number is divisible by 9, sum of its digits will also be a divisible by 9.

Step 1.
let us take a 5 digits number whose digits are A ,B ,C ,D and E(you can take any number of digits if you want...here I am taking only 5 digits...later on you will see that number of digits does not matter at all)

The number will be 10000*A + 1000 * B + 100*C + 10*D + E

Step 2. if you shuffle its digits...and let us say you get the digits in this order E,A,D,C and B( the order does not matter, you can choose any other order)
the new number is 10000*E+1000*A+100*D+10*C+B

Step 3. Subtract these you get

(10000*A-1000*A) + (1000*B-B) +(100*C-10*C) + (10*D-100*D) + (E-10000*E)

9000A+999B+90C-90D-99999E
=9*(1000*A+111*B+10*C-10*D-11111*E)
and you can see it is clearly divisible by 9


and because this is divible by 9, sum of A, B C D and E will also be divible by 9

I hope this will help...I know this is not the best way to prove it, but you will get the idea.
 
Good Arun!... s**t! I was wasting my time in different corner of the room... and its so obvious. Keep it up

-vinod


arunshamli said:
This is true for any digit number and not only for 3 or 4 digit numbers. in this particular puzzle you can start with any digit number and it will be able to give you the correct answer. aur ye jo 3digits ya 4 digits bolta hai ...woh sirf confuse karne ke liye.....

let me try to explain it


It is a well known fact that if a number is divisible by 9, sum of its digits will also be a divisible by 9.

Step 1.
let us take a 5 digits number whose digits are A ,B ,C ,D and E(you can take any number of digits if you want...here I am taking only 5 digits...later on you will see that number of digits does not matter at all)

The number will be 10000*A + 1000 * B + 100*C + 10*D + E

Step 2. if you shuffle its digits...and let us say you get the digits in this order E,A,D,C and B( the order does not matter, you can choose any other order)
the new number is 10000*E+1000*A+100*D+10*C+B

Step 3. Subtract these you get

(10000*A-1000*A) + (1000*B-B) +(100*C-10*C) + (10*D-100*D) + (E-10000*E)

9000A+999B+90C-90D-99999E
=9*(1000*A+111*B+10*C-10*D-11111*E)
and you can see it is clearly divisible by 9


and because this is divible by 9, sum of A, B C D and E will also be divible by 9

I hope this will help...I know this is not the best way to prove it, but you will get the idea.
Click to expand...
 
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