WebNote that φ ( 100) = 100 ( 1 − 1 2) ( 1 − 1 5) = 40 using Euler's product formula, so since 3 and 100 are coprime, 3 φ ( 100) ≡ 3 40 ≡ 1 (mod 10) and so 3 885 ≡ ( 3 40) 22 ⋅ 3 5 ≡ 1 22 ⋅ 3 5 ≡ 3 5 ≡ 243 ≡ 43 (mod 100), i.e. the last two digit of 3 885 are 43. Edit: I misread the question as asking only about the last digit of 3 885. Share Cite WebJul 20, 2016 · When n = 5: x = 243 The last digit is 3 There is a pattern! 1,3,9,7,1,3,9,7,1,3,9,7,1,3,9,7,1,3,9,7... and so on. The pattern repeats itself every 4 iterations. {1,3,9,7} {1,3,9,7}... We can therefore deduce that if: ( n mod 4) = 0, The last digit is 1 ( n mod 4) = 1, The last digit is 3 ( n mod 4) = 2, The last digit is 9

Find the last digit in the expression ((36472)^123!)∗((34767)^76 ...

WebThe answer is [math]60 [/math]. It can be solved using congruences modulo [math]100 [/math] but there’s an easier way. I can divide [math]234 [/math] by [math]2 [/math] and [math]345 [/math] by [math]5 [/math] to make an equivalent product as follows: [math]\quad 123\times 234\times 345\times 456\times 567\times 678\times 789 [/math] Web1. Lostsoul, this should work: number = int (10) #The variable number can also be a float or double, and I think it should still work. lastDigit = int (repr (number) [-1]) #This gives the … the cozyfeet

What is the digit in the tens place and the digit in the ... - Socratic

WebFeb 2, 2014 · finding the last digit of a number raised to another number WebAnswer (1 of 13): Last digit of 1273^n should be equal to the last digit of 3^n. Therefore, last digit of 1273^122!= 3^122! Now. 3 follows a patter in terms of its powers and last … WebMar 17, 2024 · To find last digit of a number, we use modulo operator %. When modulo divided by 10 returns its last digit. Suppose if n = 1234 then last Digit = n % 10 => 4 To find first digit of a number is little expensive than last digit. To find first digit of a number we divide the given number by 10 until number is greater than 10. the cozy tea cart milford nh