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

  • Thread starter Raza
  • Start date
In summary, the question was to compute the value of 95937 mod 2537. My friend used a calculator and discovered that the product ends with a 0, which is not possible for a number ending with 5. They then tried breaking down the problem into smaller parts and used a calculator to find the answer 1520. Another method suggested was to look at the unit group of Z/2537Z or break down Z/2537Z using the Chinese Remainder theorem. Ultimately, the correct answer was found to be 1520.
  • #1
Raza
203
0

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?
 
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  • #2
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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  • #3
removed
 
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  • #4
what is the question exactly ?
 
  • #5
how did my friend do it?
how did he break down the huge number?
'
basically
how do you compute 95937 mod 2537
 
  • #6
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/
 
  • #7
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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  • #8
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
 

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

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