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Quite easy one (self made) :)

Posted: Sun Mar 11, 2012 8:34 pm
by sm.joty
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:

Re: Quite easy one (self made) :)

Posted: Sun Mar 11, 2012 9:20 pm
by zadid xcalibured
ami coxs bazar er jonno tk jomabo vabchi.

Re: Quite easy one (self made) :)

Posted: Sun Mar 11, 2012 11:42 pm
by sm.joty
তোমার already জমে গেছে। তুমি যাত্রা শুরু কর। :D

Re: Quite easy one (self made) :)

Posted: Mon Mar 12, 2012 2:53 pm
by nafistiham
$1$
$100x+10y+z+$
$100x+10z+y+$
$100y+10x+z+$
$100y+10z+x+$
$100z+10x+y+$
$100z+10y+x=$
$222(x+y+z)=$
$37(6x+6y+6z)$
:|