I am posting the 2 ways to solve this question "

Ist way :

4¹⁰¹  = (1 + 3)¹⁰¹ = ¹⁰¹C₀ + ¹⁰¹C₁3 + ¹⁰¹C₂3² + ¹⁰¹C₃3³ + ............+¹⁰¹C₁₀₁3¹⁰¹

                      4¹⁰¹  = 1 + ¹⁰¹C₁3 + ¹⁰¹C₂3² + ¹⁰¹C₃3³ + .....................+3¹⁰¹                                

we know ⁿC_r is always an integer and each term conatins a multiple of 101 apart from 1st and last one.

so   4¹⁰¹ when divided by 101 gives the same remainder what 1 + 3¹⁰¹ divided by 101 will give.

so,

1 + 3¹⁰¹ = 1 + (1 + 2)¹⁰¹ = 1 + ¹⁰¹C₀ + ¹⁰¹C₁2 + ¹⁰¹C₂2² + ¹⁰¹C₃2³ + ............+¹⁰¹C₁₀₁2¹⁰¹

                            1 + 3¹⁰¹ = 1 + 1 + ¹⁰¹C₁2 + ¹⁰¹C₂2² + ¹⁰¹C₃2³ + ............+2¹⁰¹

same logic goes here as well,

as every term conatins a multiple of 101 apart from 1st and last one.

so 1 + 3¹⁰¹ when divided by 101 will give the same remainder what 1 + 1 + 2¹⁰¹ will give.

2 + 2¹⁰¹ = 2 + (1 + 1)¹⁰¹ = 2 + ¹⁰¹C₀ + ¹⁰¹C₁ + ¹⁰¹C₂ + ¹⁰¹C₃ + ............+¹⁰¹C₁₀₁

                                            = 2 + 1 + ¹⁰¹C₁ + ¹⁰¹C₂ + ¹⁰¹C₃ + ............ + 1

                                            = 4 + ¹⁰¹C₁ + ¹⁰¹C₂ + ¹⁰¹C₃ + ............

this when divided by 101 will give 4 as the remainder

as every other term contains 101.

Ans = 4

2nd way :

From Fermat's Little theorem, we know the following result

If a is an integer and p is a prim number, then a^p - a is divisible by p

Here a = 4 (an integer)

          p = 101 (a  prime number)

hence according to theorem , 4¹⁰¹ - 4 is divisible by 101

i.e. when 4¹⁰¹ is divided by 101, it gives 4 as the remainder.

4¹⁰¹ - 4 = 101k     where k is a positive integer

 4¹⁰¹ = 101k +  4    i.e. 4 is the remainder.   Ans

But, try to remember Fermat's Little Theorem, it's of good help.