Lets start with the remainder, since that will be a pretty straightforward exercise in modular arithmetic. 
The big number can be written as 4*10^101 + 1, and we want this mod 41. 
4*10^101 + 1 

= 2^103 * 5^101 + 1 

= 2^4*(2^7)^14 * 5^2*(5^3)^33 + 1.

2^7=128=5 mod 41 and 5^3=125=2. So working in mod 41: 

2^4*(2^7)^14 * 5^2*(5^3)^33 + 1 

= 2^4*5^14 * 5^2*2^33 + 1

= 2^37 * 5^16 + 1

= 2^2*(2^7)^5 * 5*(5^3)^5 + 1

= 2^2*5^5 * 5*2^5 + 1 

= 2^7 * 5^6 + 1

= 2^7 * (5^3)^2 + 1

= 5 * 2^2 + 1

= 41 = 0 mod 41.

The remainder is 0. So our big number is a multiple of 41.  Onto the quotient.

4*10^101 + 1 = 41*10^100 - 999....999 (a number composed of 100 9's)

41*10^100 is obviously a multiple of 41 and because 999....999 (100 9's) is the difference of two multiples of 41, it is also a multiple of 41.

By Fermat's Little Theorem we know 10^40 = 1 mod 41, thus 99...99 (40 9's) is a multiple of 41.  But there may be a smaller number of 9's that is divisible by 41.  Fermat's Little Theorem also tells us that number of 9's is a factor of 40.

So we have candidates starting with 9, 99, 9999, 99999, 99999999 etc.  Fortunately the fourth candidate gives us 99999/41 = 2439.

Then 999....999 (100 9's) divided by 41 will be 20 blocks of 02439 concantenated.

Then (4*10^101 + 1)/41 

= (41*10^100 - 999....999)/41 

= 1 + 99999...99999 - 0243902439....02439 (20 blocks each of 99999 and 02439)

= 1 + 97560....97560 (20 blocks)

= 97560....9756097561 (19 blocks of 97560 followed by 97561)

This is our quotient.