Given a positive N, the task is to find the last two digits of 7N.
Input: N = 5
The value of 75 = 7 * 7 * 7 * 7 * 7 = 8507
Therefore, the last two digits are 07.
Input: N = 12
The value of 712 = 13841287201
Therefore, the last two digits are 01.
Approach: A general approach to finding the last K digits of XY is to discuss this article in logarithmic time complexity. In this article, we will discuss the constant time solution.
Below is the observation for the value of 7N for some values of N:
71 = 7 last two digit = 07
72 = 49 last two digit = 49
73 = 243 last two digit = 43
74 = 2401 lasr two digit = 01
75 = 16807 last two digit = 07
76 = 117649 last two digit = 49
77 = 823543 last two digit = 43
78 = 5764801 last two digit = 01
Based on the above observations we have the following cases:
- If the last two digit in 7N = 07 when N = 4K + 3.
- If the last two digit in 7N = 49 when N = 4K + 2.
- If the last two digit in 7N = 43 when N = 4K + 1.
- If the last two digit in 7N = 01 when N = 4K.
Below is the implementation of the above approach:
Time Complexity: O(1)
Auxiliary Space: O(1)
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