Rare Numbers

                   "The numbers, which gives a perfect square on adding as well as subtracting its reverse are rare and hence termed as Rare Numbers."


                             If "X" is a positive integer and "X_rev" is the integer obtained from X by writing its decimal digits in reverse order, then if X + X_rev and X - X_rev both are perfect squares then X is termed as Rare Number.
                             Shyam Sunder Gupta (from INDIA) have been investigating these numbers since 1989. He has developed computer program in Fortran to compute Rare numbers.

Examples : 65 and 621770

What will be the remainder when 101th power of 4 is divided by 101 ?

What will be the remainder 4¹⁰¹ is divided by 101 ?
A. 1
B. 2
C. 3
D. 4

Solution : 

Ist Method :

From Binomial Theorem
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 except 1st and last one.
so   4¹⁰¹ when divided by 101 gives the same remainder what 1 + 3¹⁰¹ divided by 101 will give.



Now,
1 + 3¹⁰¹ = 1 + (1 + 2)¹⁰¹ = 1 + ¹⁰¹C₀ + ¹⁰¹C₁2 + ¹⁰¹C₂2² + ¹⁰¹C₃2³ + ............+¹⁰¹C₁₀₁2¹⁰¹
                            1 + 3¹⁰¹ = 1 + 1 + ¹⁰¹C₁2 + ¹⁰¹C₂2² + ¹⁰¹C₃2³ + ............+2¹⁰¹




As every term conatins a multiple of 101 except 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 Method :

From Fermat's Little theorem, we know the following result
                         If a is an integer and p is a prime 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 = 101 k     where k is a positive integer
                        4¹⁰¹ = 101 k +  4  
                 i.e. 4 is the remainder.