Program to find last digit of the given sequence for given n in Python

Suppose we have a value n. We have to find the last digit of sequence S. The equation of S is given below −

$$\sum_{i=0\: 2^{^{i}}\leqslant n}^{\alpha } \sum_{j=0}^{n} 2^{2^{^{i}+2j}}$$

So, if the input is like n = 2, then the output will be 6 because: here only i = 0 and i are valid, so

• S₀ = 2^(2^0 + 0) + 2^(2^0 + 2) + 2^(2^0 + 4) = 42
• S₁ = 2^(2^1 + 0) + 2^(2^1 + 2) + 2^(2^1 + 4) = 84 The sum is 42+84 = 126, so last digit is 6.

To solve this, we will follow these steps −

• total:= 0
• temp := 1
• while temp <= n, do
  • total := total + (2^temp mod 10)
    • temp := temp * 2
  • total := total * (1 +(4 when n is odd otherwise 0)) mod 10
• return total

Example

Let us see the following implementation to get better understanding −

def solve(n):
   total= 0
   temp = 1
   while (temp <= n):
      total += pow(2, temp, 10)
      temp *= 2
   total = total * (1 + (4 if n %2 ==1 else 0)) % 10
   return total

n = 2
print(solve(n))

Input

2

Output

6