Help Needed: Solving 2007^2007^2007^2007 Last 3 Digits

  • Thread starter AraProdieur
  • Start date
  • Tags
    Assistance
The trick is to use Euler's theorem: a^phi(n) = 1 mod n, when a is relatively prime to n. We know that 2007 = 3*11*61. The number we are interested in is: 2007^2007^2007^2007 mod 1000. We can use the fact that 2007 = 61*33, and that 61 is relatively prime to 1000. By Euler's theorem, 61^phi(1000) = 1 mod 1000. So, if we can find phi(1000), that will give us a way to reduce the exponent of 2007 mod 1000
  • #1
AraProdieur
27
0
What are the last three digits of 2007^2007^2007^2007?

So far, I have tried doing it one by one to see if there was a pattern in the units, but haven't had much luck.

Any help?
 
Physics news on Phys.org
  • #2
Perhaps think of it in a different manner: what is the remainder of 2007^n when divided by 1000 ?
 
  • #3
This is a pretty tricky problem in modular arithemetic. Here's a simpler problem of the same sort: what is the remainder when 2^1000 is divided by 13 ?

Solution: notice that 64 = -1 mod 13, since 5*13 = 65. Also, 64 is a power of 2: 2^6 = 64. Then, 1000 = 6*166 + 4, so 2^1000 = (2^6)^166 * 2^4 = 64^166 * 2^4. Then, 2^1000 mod 13 = (64^166 mod 13)*(2^4 mod 13) = (64 mod 13)^166 * (16 mod 13) = (-1)^166 * 3 = 3. So the answer is 3.

Your problem is considerably more difficult.
 

1. What is the problem of "Help Needed: Solving 2007^2007^2007^2007 Last 3 Digits"?

The problem is to find the last three digits of the number 2007 raised to itself, repeated 2007 times.

2. Is there a straightforward solution to this problem?

No, there is not a straightforward solution. This problem requires advanced mathematical techniques and computations.

3. Why is this problem important?

This problem is important because it challenges our understanding of large numbers and their properties. It also has practical applications in fields such as cryptography and number theory.

4. What is the current progress in solving this problem?

There is currently no known solution to this problem. However, mathematicians have made progress in finding the last digit of the number 2007 raised to itself, repeated a certain number of times.

5. Is there an efficient way to solve this problem?

No, there is not an efficient way to solve this problem. The number of calculations required to solve this problem increases exponentially with each repetition of 2007 raised to itself. It may require powerful computers and advanced algorithms to find a solution.

Similar threads

  • Calculus and Beyond Homework Help
Replies
1
Views
81
  • Special and General Relativity
Replies
6
Views
4K
  • STEM Academic Advising
Replies
1
Views
179
  • Calculus and Beyond Homework Help
Replies
4
Views
2K
  • Calculus and Beyond Homework Help
Replies
8
Views
891
Replies
17
Views
2K
Replies
72
Views
5K
  • Introductory Physics Homework Help
Replies
7
Views
139
  • Calculus and Beyond Homework Help
Replies
32
Views
2K
  • Computing and Technology
Replies
8
Views
12K
Back
Top