An integer $N$ is formed by writing first 2012 natural numbers left to right.
$N=1234567891011.....20112012$
How many $2$'s are there in $N$ ?
How many $0$'s are there in $N$ ?
Find the remainder when $N$ is divided by $9$ .
N(1234...20112012)
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sakibtanvir
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An amount of certain opposition is a great help to a man.Kites rise against,not with,the wind.
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MATHPRITOM
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Re: N(1234...20112012)
I give answer of first question.WE can think about the problem in a way that we want to count the number of 2's form 1 to 2002.
when , once 2 will be considered in a number a time, it will not be considered again. we will think for the numbers where 2 is presented more than one time later.
a)First, consider 1 digits number. we can get it by only 1 ways.
b)now, consider 2 digit numbers. let, 2 is in first position(একক স্থানীয় ঘর ). so,here the number of choose is 8.
c) now, consider 2 digit number where 2 is in 2nd position (দশক স্থানীয় ঘর ) .so, here the number of choose is 9.
d) 3rd digit number & 2 is in 1st position ,so, the possibilities are 9*8.
e) 3rd digit number & 2 is in 2nd position . again, possible condition is 9*8.
f) 3rd digit number & 2 is in 3rd position. so, possible condition is 9*9.
g) 4 digit number & 1 is in 4th position & 2 is in 1st,2nd or 3rd position .so, possible condition 3*9*9.
h)now, remaining numbers after the range 2000,2001,2002,...,2012. 2 is contained 14 times by them .
now, we consider the case where 2 appears more than 1 time.
i) for 2 digit numbers , 1 case.
j) for 3 digits , all digits are 2 is 1 case. for 2 digits are 2 the case is 9+9+8.
k) for 4 digits within limit 1000 to 1999, there are 4 possible conditions ..it's 3*9+1.
yahoo.. now, the answer is a+b+c+d+...+k.
when , once 2 will be considered in a number a time, it will not be considered again. we will think for the numbers where 2 is presented more than one time later.
a)First, consider 1 digits number. we can get it by only 1 ways.
b)now, consider 2 digit numbers. let, 2 is in first position(একক স্থানীয় ঘর ). so,here the number of choose is 8.
c) now, consider 2 digit number where 2 is in 2nd position (দশক স্থানীয় ঘর ) .so, here the number of choose is 9.
d) 3rd digit number & 2 is in 1st position ,so, the possibilities are 9*8.
e) 3rd digit number & 2 is in 2nd position . again, possible condition is 9*8.
f) 3rd digit number & 2 is in 3rd position. so, possible condition is 9*9.
g) 4 digit number & 1 is in 4th position & 2 is in 1st,2nd or 3rd position .so, possible condition 3*9*9.
h)now, remaining numbers after the range 2000,2001,2002,...,2012. 2 is contained 14 times by them .
now, we consider the case where 2 appears more than 1 time.
i) for 2 digit numbers , 1 case.
j) for 3 digits , all digits are 2 is 1 case. for 2 digits are 2 the case is 9+9+8.
k) for 4 digits within limit 1000 to 1999, there are 4 possible conditions ..it's 3*9+1.
yahoo.. now, the answer is a+b+c+d+...+k.
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sakibtanvir
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Re: N(1234...20112012)
I solved it like you.But it kills a lot of time 
.I wanted a combinatorial solution of this.
.I wanted a combinatorial solution of this.An amount of certain opposition is a great help to a man.Kites rise against,not with,the wind.
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nafistiham
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Re: N(1234...20112012)
well, I don't think there is any other way but counting.sakibtanvir wrote:I solved it like you.But it kills a lot of time
\[\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0\]
Using $L^AT_EX$ and following the rules of the forum are very easy but really important, too.Please co-operate.
Using $L^AT_EX$ and following the rules of the forum are very easy but really important, too.Please co-operate.
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MATHPRITOM
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Re: N(1234...20112012)
There is a quote in a book about combinatorics " Often, counting the cases of the problem is also a great combinatorial problem." I think, this is such a problem. হা হা হা ......।।