WebJan 2, 2015 · 1 Answer. Sorted by: 1. if it's a string you want the last two characters, Else if it;s an integer doing %100 first storing the result in an int and then doing %10 on that is about as efficient as it gets. ( short might work better than int here, might not) integers are stored as binary numbers and the CPU can't see the 'last base 10 digit ... WebJun 18, 2024 · 8. Problem is easy to solve once you realize that the last digits of powers form a cycle. For example: 2: 2, 4, 8, 6, 2, 4 3: 3, 9, 7, 1, 3, 9. With that in mind you can create the cycle first and then just index it with modulo of n2: def last_digit (n1, n2): if n2 == 0: return 1 cycle = [n1 % 10] while True: nxt = (cycle [-1] * n1) % 10 if nxt ...
What is the last two digits of 65*29*37*63*71*87*62
Web42 166 = (2 × 21) 166. The last two digits of 76 n where n is a positive integer will be 76. Also, for numbers with the last two digits as 24, the square of the number will have 76 as the last digit. 2 10 ends with 24, ⇒ 2 166 21 166 = (2 10) 16 × 2 6 × 21 166. The 10s place will be the unit digit of the result of the multiplication of 10s ... WebLet (87) 75 63 55 = (87) xwhere x = (75) 63 55 = (87) odd∵ Cyclicity of 7 is 4∴ To find the last digit we have to find the remainder when x is divided by 4x = (75) oddpower = (76−1) oddpowerWhere x is divided by 4 so remainder will be -1 or 3 but remainder should be always positiveTherefore the last digit of (87) 75 63 55 will be (7) 3 ... bucher conjugaison
c - Finding the last 10 digits of 2^n - Stack Overflow
WebFind the last two digits of 87 474. 87 474 = 87 472 × 87 2 = (87 4) 118 × 87 2 = (69 × 69) 118 × 69 (The last two digits of 87 2 are 69) = 61 118 × 69 = 81 × 69 = 89. If you … WebThe last digit of 32^33 is equal to last digit of 2^33. Because only last digit matters for multiplying, for calculating the last digit of answer. And now, you can calculate 2^33 by … WebSo you want to access the digits in a integer like elements in a list; easiest way I can think of is: n = 56789 lastdigit = int (repr (n) [-1]) # > 9 Convert n into a string, accessing last element then use int constructor to convert back into integer. For a Floating point number: bucher contemporain