Spletthe question can be written as (2^67)* (7^49). So first we find the last digit of 2^67 and last digit of 7^49. Then the product of those both digits. For finding the last digit of 2^67, there must be a pattern involved. So, 2^1=2 2^2=4 2^3=8 2^4=16 2^5=32 2^6=64 ... ... It is clear that the last digit repeats at a difference of 4 in the power of 2. Splet27. apr. 2024 · The last digit in 2^11 = 2048 is 8 A Naive Solution is to first compute power = pow (2, n), then find the last digit in power using power % 10. This solution is inefficient and also has an integer arithmetic issue for slightly large n. An Efficient Solution is based on the fact that the last digits repeat in cycles of 4 if we leave 2^0 which is 1.
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Spletpred toliko urami: 3 · Oilers. -175. -220. -1.5 (-140) The game and series odds are comparable to last season's meeting, and the Kings can be found for as high as +210 in … SpletYou are given two integer numbers: the base a (0 <= a <= 20) and the index b (0 <= b <= 2,147,483,000), a and b both are not 0. You have to find the last digit of a b. Input The first line of input contains an integer t, the number of test cases (t <= 30). t test cases follow. For each test case will appear a and b separated by space. Output Splet- Addition: 2-digit plus a 2-digit number up a total of 30, 50 or 100 - Subtraction: Two 2-digit numbers with the first number a maximum of 30, 50 or 100 - Multiplication: From 2 x 2 to 12 x 12 - Division: Dividend a maximum of 100, Divisor up to 12. The work is broken down into modules so that a step by step approach can be taken. slaughterhouse auto