Finding Last Digits of 3400: Without Calculator or Euler's Totient Function

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Homework Help Overview

The discussion revolves around finding the last digits of the expression 3400, specifically focusing on the last one, two, and three digits. Participants explore methods to achieve this without the use of a calculator, considering concepts such as modular arithmetic and Euler's totient function.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the necessity of calculating 3400 mod 10, 100, and 1000, and whether Euler's totient function is required. Some suggest that finding 3^400 mod 1000 could be a straightforward approach. Questions arise about the implications of results obtained through Euler's theorem and how to handle other numbers like 5623.

Discussion Status

The conversation is active, with participants offering various insights and methods. Some express uncertainty about the application of Euler's theorem and the behavior of powers of 5 mod 1000. There is an exploration of patterns in powers of 5, with some participants questioning the mathematical reasoning behind observed results.

Contextual Notes

Participants note the challenge of applying Euler's theorem and the implications of specific calculations, such as the gcd of certain numbers. There is also mention of the differences in behavior of powers of 5 under different moduli, indicating a need for further exploration of these concepts.

mcooper
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Homework Statement



How can I find the last 1,2 and 3 digits of 3400?

The Attempt at a Solution



I know I have to find 3400 mod 10, 100 and 1000 but I want to know how to do this without a calculator. Do I have to use Euler's totient function?

thanks in adavnce
 
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All you need is 3^(400) mod 1000. That will give you the last three digits. Sure, use the totient function and Euler's theorem. I think it's the easiest way.
 
Thanks a lot! I am just wondering, when I get 3400= 1 mod 1000 by Euler's Theorem does that mean the last 3 digits are 001?

Also I kind of cheated a bit because I knew that phi(1000)=400. What can I use if I have a "random" number to calculate such as 5623? Would I just systematically go through, maybe using the the totient function and Euler's theorem, until I get something manageable?
 
Sure, if a number is equal to 1 mod 1000 then it's last three digits are 001. And 3^400 mod 1000 is a particularly easy case. 5^623 mod 1000 is harder. You could start with 5^4 mod 1000=625. Then to get 5^8 mod 1000, you square the 625 and take the result mod 1000. Then you can square again to get 5^16 mod 1000. You can get the exponent pretty high pretty fast. But it turns out the powers of 5 mod 1000 are pretty easy for another reason. Can you find it?
 
Last edited:
Are the powers of 5 easy because (5,1000)=1 so you can use Euler's Theorem?
 
mcooper said:
Are the powers of 5 easy because (5,1000)=1 so you can use Euler's Theorem?

How do you figure gcd(5,1000)=1? You can't apply Euler's theorem. Just start looking at increasing powers of 5 mod 1000. There's a simple pattern.
 
Sorry that's poor, its been a long day/night. All powers of 5 are congruent to 625 mod 1000 (hence 25 mod 100). Kind of weird but kind of cool. Is there a mathematical reason/proof for this? mod 10000 they are not all the same...
 
mcooper said:
Sorry that's poor, its been a long day/night. All powers of 5 are congruent to 625 mod 1000 (hence 25 mod 100). Kind of weird but kind of cool. Is there a mathematical reason/proof for this? mod 10000 they are not all the same...

That's not true. 5^4 is congruent to 125. 5^5 is conguent to 625. It's just because 125*5 is congruent to 625 and 625*5 is congruent to 125. They keep repeating.
 

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