NMTC 2011 PROBLEM

try 3 digit number $101 \times 2011 = 203111$

the sum of the digits of the result is $8$

try 5 digit number $10101 \times 2011 = 20313111$

the sum of the digits of the result is $12$

try 7 digit number $1010101 \times 2011 = 2031313111$

the sum of the digits of the result is $16$

so we have three coordinates $(3,8) , (5,12) , (7,16)$

so that $x =$ number of digits , $y$ = the sum of them

we notice that it follows this function

$f(x) = 2x+2$

that will lead to

$f(2011) = 2*2011+2 =4024$

For a 3 digit number 101 x 2011= 203111

For a 5 digit number 10101 x 2011= 20313111

For a 7 digit number 1010101 x 2011= 2031313111

Notice that the number of 3's in the product is always equal to the number of 0's, the number of 1's in the product is always one more than the number of 1's in the digit and there's always one 2.

Therefore, for a 2011 digit number (which has 1006 ones and 1005 zeroes), the sum of product would be =

3 x 1005 + 1007 x 1 + 2= 4024

$We\quad would\quad obsereve\quad a\quad series\quad of\quad sum\quad as\quad follows\\ 2011\times 1=4\quad \quad (sum\quad of\quad digits)\\ 2011\times 101=8\\ 2011\times 10101=12\\ .\\ .\\ .\\ 2011\times 1010.....upto\quad n\quad times\quad '1'=4+(n-1)4\\ therefore\quad a\quad 2011\quad digit\quad no.\quad would\quad have\quad 1006\quad '1'\\ therefore\quad 4+(1005)4=4024\\ \\ By\quad Mehul$

$\boxed{4024}$