Solving Modular Exponentiation: How Did My Friend Break Down 95937 mod 2537?

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Homework Help Overview

The problem involves calculating 95937 mod 2537, with participants discussing various approaches and breakdowns of the computation.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Some participants describe attempts to break down the large exponentiation and question the validity of the methods used, including the use of calculators and modular properties.
  • Others suggest exploring the unit group of Z/2537Z and the Chinese Remainder Theorem as potential avenues for simplifying the problem.
  • There are inquiries about the exact nature of the question and how to compute the modular exponentiation manually.

Discussion Status

The discussion is ongoing, with various interpretations and methods being explored. Some participants have shared results obtained through software, while others are focused on manual calculations and theoretical approaches.

Contextual Notes

Participants note potential confusion regarding the setup of the problem and the assumptions made about the calculations. There is also mention of discrepancies in results obtained through different methods.

Raza
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Homework Statement



95937 mod 2537

Homework Equations


None


The Attempt at a Solution


Here's what my friend did:
Clipboard01-1.jpg

How do I do this? and how did he do this?
 
Last edited:
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Raza said:
[tex]95 \times 1786 \times 341 \times 2188 \times 403[/tex]

How do I do this? and how did he do this?
This can't be right, because the product above ends with a 0. Powers of any number that ends with a 5 will also end with a 5 (except if the exponent is 0). In fact, for n >=5,
95n ends with "09375" if n is odd, and
95n ends with "90625" if n is even.

If found this through using Calculator in Windows 7. :wink:

Furthermore,
Raza said:
Here's what my friend did:
[tex]95^{937}[/tex]

[tex]95 \times 240^{234}[/tex]
This isn't right either. According to Calculator,
95937 ≈ 1.33 * 101853, but
95 * 240234 ≈ 8.85 * 10558.
(This assumes that there is no overflow error here.)

In fact, 95937 should equal 95 * (81450625)234.

Is the entire problem? Perhaps it should be 95937 mod some number?
 
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removed
 
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what is the question exactly ?
 
how did my friend do it?
how did he break down the huge number?
'
basically
how do you compute 95937 mod 2537
 
I see, well the answer is 1520 but I found that by using MAGMA. Do not know yet how to do it manually.

Code:
n := 2537;
R := ResidueClassRing(n);

number := R ! 95; 
number ^ 937;

http://magma.maths.usyd.edu.au/calc/
 
What you could do is looking at the unit group of Z/2537Z. I found out it is isomorphic to Z/2Z x Z/1218Z however I used MAGMA too for this one.

Another trick (it seems your friend did something like that): break down Z/2537Z. You have (at least when it comes to addition) an isomorphism with Z/pZ x Z/qZ (do not confuse this with the isomorphism of the unit group). You can repeat that, it has to do with the Chinese Remainder theorem I think.
 
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Okay, I had this question in my exam and after being verified by other people, I think I got it correct.
95937 mod 2537 = 95512 mod 2537 x 95256 mod 2537 x 95128 mod 2537 x 9532 mod 2537 x 958 mod 2537 x 951 mod 2537

So now you do the mod from the power of 1 at .
951 mod 2537 = 95
952 mod 2537 = 1414
954 mod 2537 = (952)2 mod 2537 = 14142 mod 2537 = 240
958 mod 2537 = (954)2 mod 2537 = 2402 mod 2537 = 1786
9516 mod 2537 = (958)2 mod 2537 = 17862 mod 2537 = 787
9532 mod 2537 = (9516)2 mod 2537 = 7872 mod 2537 = 341
9564 mod 2537 = (9532)2 mod 2537 = 3412 mod 2537 = 2116
95128 mod 2537 = (9564)2 mod 2537 = 21162 mod 2537 = 2188
95256 mod 2537 = (95128)2 mod 2537 = 21882 mod 2537 = 25
95512 mod 2537 = (95256)2 mod 2537 = 252 mod 2537 = 625

From the first equation:
95937 mod 2537 = (625 x 25 x 2188 x 341 x 1786 x 95) mod 2537
95937 mod 2537 = (625 x 25) mod 2537 x (2188 x 341) mod 2537 x (1786 x 95) mod 2537
95937 mod 2537 = (403 x 230 x 2228) mod 2537 = 1520
 

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