Question

What are the digits in the units and ten's place for this number, 74^2563

Answer 1

To find the digits in the units place, we need to find \[74^{2563} \mod 10\]Which is equivalent to \[4^3 \mod 10\]With some quick calculations, the last digit is 4.

Answer 2

in the ten's place?

Answer 3

Also, how did you get 74^2563 % 10 as 4^3 Mod 10?

Answer 4

To find the ten's place, we need to find \[74^{2563} \mod 100\]Otherwise known as \[74^{63} \mod 100\]Using Euler's Theorem, we know that \(\phi(100)=40\) So we need to find \[74^{40+23}\equiv 74^{23} \mod 100\]Here, we can use successive squaring (or Wolfram) to finish it off. Thus we get that the tens digit is 2.

Answer 5

As to the question you just asked, I was able to reduce to \(4^3 \mod 10\) because we're only focusing on the last digit. We were able to just throw away the other digits.

Answer 6

For the first one cyclicity trick is a better choice.

Answer 7

I think Euler's Theorem is not applicable here as \( (74,100) \neq 1 \)

Answer 8

Isn't 100 and 74 not co-prime?

Answer 9

You are correct. My apologies.

Answer 10

I am using Binomial theorem, the tens digit is 2.

Answer 11

the funny thing is, using eulers theorem @KingGeorge got the last digit as 4 which actually is.

Answer 12

I didn't use Euler's for the last digit. It was small enough to just multiply out. Also, \[74^{63}\equiv74^{23 } \mod 100\]But not by Euler's theorem. It would take an application of the Chinese Remainder theorem.