Quite easy one (self made) :)

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sm.joty
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Quite easy one (self made) :)

Unread post by sm.joty » Sun Mar 11, 2012 8:34 pm

1. We have a three digit number where no of those digit is zero and the digits are individual (that means no repetition of digit). Prove that all permutation of the number is divisible by $37$.
For instance: Let the number is $123$, then sum of all permutation of $123$ is $1332$

2.If $A=[0,1,2,........,9]$ then we have a $n$ digit number formed by using the digits from $A$. The sum of all the possible numbers is $S$. Then prove that, $S$ has $2n$ digit for all $n \in \mathbb{N}$ and $n\geq 2$
Note that here you're free for repetition. But you can't use zero for the first digit.

3. Assume that some of the numbers are wiped out form set $A$. Then what is the generalization for finding S and the number of digit of $S$.

N.B:This problems are a little manipulation of a problem from MILON da. Thanks to MILON da. :mrgreen:
হার জিত চিরদিন থাকবেই
তবুও এগিয়ে যেতে হবে.........
বাধা-বিঘ্ন না পেরিয়ে
বড় হয়েছে কে কবে.........

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zadid xcalibured
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Re: Quite easy one (self made) :)

Unread post by zadid xcalibured » Sun Mar 11, 2012 9:20 pm

ami coxs bazar er jonno tk jomabo vabchi.

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sm.joty
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Re: Quite easy one (self made) :)

Unread post by sm.joty » Sun Mar 11, 2012 11:42 pm

তোমার already জমে গেছে। তুমি যাত্রা শুরু কর। :D
হার জিত চিরদিন থাকবেই
তবুও এগিয়ে যেতে হবে.........
বাধা-বিঘ্ন না পেরিয়ে
বড় হয়েছে কে কবে.........

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nafistiham
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Re: Quite easy one (self made) :)

Unread post by nafistiham » Mon Mar 12, 2012 2:53 pm

$1$
$100x+10y+z+$
$100x+10z+y+$
$100y+10x+z+$
$100y+10z+x+$
$100z+10x+y+$
$100z+10y+x=$
$222(x+y+z)=$
$37(6x+6y+6z)$
:|
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