Gauss 10 Crack

On this page we look at the Chinese Remainder Theorem (CRT), Gauss's algorithm to solve simultaneous linear congruences, a simpler method to solve congruences for small moduli, and an application of the theorem to break the RSA algorithm when someone sends the same encrypted message to three different recipients usingthe same exponent of e=3.

However, these Gauss-points will just be the ones that are already seriously damaged and where extrapolation errors will be rather negligible. When, for example, the 10% percentile is considered, the value of the global cycle jump NJUMP would be 4078 cycles for the example in Figure 2.17. This is usually almost 10 times stronger than the surface Gauss. Therefore a magnet with 1,500 surfaces Gauss will have about 14,000 core Gauss. In order for Biomagnetism to function correctly, magnets of at least 1,000 surface Gauss or 11,000 Core Gauss or BrMax Gauss or above ratings are desired for optimal use. VG Virgo Cracks APTECH Gauss see 6. 25 crack someone of them may contain popups and banners, all of which are known and have APTECH Gauss 6. 25 crack or keygen download. When you still have trouble downloading APTECH gauss or another file, please add it in the comments below and our support team or a community member, helfen. Oct 23, 2010 On this page we look at the Chinese Remainder Theorem (CRT), Gauss's algorithm to solve simultaneous linear congruences, a simpler method to solve congruences for small moduli, and an application of the theorem to break the RSA algorithm when someone sends the same encrypted message to three different recipients using the same exponent of e=3. The GAUSS prompt ‘’ will show up on the screen waiting for us to type GAUSS commands. In GAUSS terminology, the GAUSS environment we are in is referred to as the GAUSS command mode. We can directly type GAUSS statements on the screen and execute them (a procedure that is called running.

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The Chinese Remainder Theorem

Theorem. Let $n_1,n_2,ldots,n_r$ be positive integers such that $gcd(n_i,n_j)=1$ for $ine j$.Then the system of linear congruencesbegin{align}x & equiv c_1 pmod{n_1} x & equiv c_2 pmod{n_2} x & equiv c_3 pmod{n_3} & ldots x & equiv c_r pmod{n_r}end{align}has a simultaneous solution which is unique modulo $n_1n_2cdots n_r$.
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Note that all the theorem says is that there is a unique solution. It doesn't actually say how to solve it. This is usually done using Gauss's algorithm.There is also a variant of the CRT used to speed up the calculations in the RSA algorithm.

The name 'Chinese' comes from an old Chinese puzzle allegedly posed by Sun Tsu Suan-Ching in 4 AD:

There are certain things whose number is unknown. Repeatedly divided by 3, the remainder is 2; by 5 the remainder is 3; and by 7 the remainder is 2. What will be the number?

In modern number theory, we would write that as a problem to solve the simultaneous congruences

The Chinese Remainder Theorem (CRT) tells us that since 3, 5 and 7 are coprime in pairs then there is a unique solutionmodulo 3 x 5 x 7 = 105. The solution is x = 23. You can check that by noting that the relations

are all satisfied for this value of x.

Gauss's algorithm

Algorithm. Let N=n1n2...nr then

where Ni = N/ni and di ≡ Ni-1 (mod ni).

The latter modular inverse di is easily calculated by the extended Euclidean algorithm.You can also use the bd_modinv utility in ourModular Arithmetic Freeware download.

Example

For the original 'Chinese' problem above we have

n1=3, n2=5, n3=7
N = n1n2n3 = 3 x 5 x 7 = 105
c1=2, c2=3, c3=2.

Now N1 = N/n1 = 35 and so d1 = 35-1 (mod 3) = 2,
N2 = N/n2 = 21 and so d2 = 21-1 (mod 5) = 1, and
N3 = N/n3 = 15 and so d3 = 15-1 (mod 7) = 1. Hence

Another example

Using Gauss's algorithm,

n1=3, n2=4, n3=5
N = n1n2n3 = 3 x 4 x 5 = 60
c1=1, c2=2, c3=3.
N1 = N/n1 = 20; d1 = 20-1 (mod 3) = 2 [check: 2x20=40≡1 (mod 3)]
N2 = N/n2 = 15; d2 = 15-1 (mod 4) = 3 [check: 3x15=45≡1 (mod 4)]
N3 = N/n3 = 12; d3 = 12-1 (mod 5) = 3 [check: 3x12=36≡1 (mod 5)]
x ≡ c1N1d1 + c2N2d2 + c3N3d3 (mod N)
x = (1x20x2) + (2x15x3) + (3x12x3) = 238 ≡ 58 (mod 60)

so a solution is x = 58.Note that this is 'a' solution. Any integer that satisfies 58 + 60k for any integer k is also a solution, but the method gives you the unique solution in the range 0 ≤ x < n1n2n3.

A simpler method

For congruences with small moduli there is a simpler method (useful in exams!).To solve the previous problem, write out the numbers x ≡ 3 (mod 5) until you find a number congruent to 2 (mod 4),then increase that number by multiples of 5 x 4 until you find number congruent to 1 (mod 3).

We find it easier to start with the largest modulus and work downwards.

To solve the original Chinese problem:

Cracking RSA

Alice sends the same message m encrypted using the RSA algorithm to three recipients with different moduli n1,n2,n3 all coprime to each other but using the same exponent e=3. Eve recovers the three ciphertext values c1,c2,c3 and knows the public keys (n,e=3) of all the recipients. Can Eve recover the message without factoring the moduli?

Yes. Eve uses Gauss's algorithm above to find a solution x, in the range 0 ≤ x < n1n2n3,to the three simultaneous congruences

We know from the Chinese Remainder Theorem that m3 < n1n2n3, so it follows that x = m3 and so m can be recovered by simplycomputing the integer cube root of x. Note that the cube root does not involve any modular arithmetic and so is straightforward to compute (well, as straightforward as computing any cube root is).

Example

There are three recipients with public keys (87,3), (115,3) and (187,3).That is, we have e=3 and

n1=29x3=87, n2=23x5=115, n3=17x11=187

(although the factorisation would neither be public nor feasibly computable for large n's used in practice)

Alice encrypts the message m=10 using RSA to all three, as follows,

c1 = 103 mod 87 = 43; c2 = 103 mod 115 = 80; c3 = 103 mod 187 = 65

and these three ciphertext values c1, c2, c3 are intercepted by Eve,who also knows the public values (ni, e).She then uses Gauss's algorithm as follows

N = n1n2n3 = 87 x 115 x 187 = 1870935
N1 = N/n1 = 115x187 = 21505; d1 = 21505-1 (mod 87) = 49
N2 = N/n2 = 87x187 = 16269; d2 = 16269-1 (mod 115) = 49
N3 = N/n3 = 87x115 = 10005; d3 = 10005-1 (mod 187) = 2
x ≡ c1N1d1 + c2N2d2 + c3N3d3 (mod N)
x = (43.21505.49) + (80.16269.49) + (65.10005.2) = 110386165 ≡ 1000 (mod 1870935)

So m is the cube root of 1000; that is, m = 10, as required.Eve did not need to factor the moduli to find the message.

To compute the modular inverses, we used the bd_modinv function in our Modular Arithmetic Freeware package (new updated version released 11-11-11)

Comment

In practice with RSA we would be looking at much larger moduli in the order of 1000 or 2000 bits (i.e. numbers about 300 to 600 decimal digits long, probably too big for your pocket calculator), but the same principles apply.You would need to use a computer package that does large integer arithmetic (like our free BigDigits software - see below).It is most likely that any three moduli in practice will be coprime, so the method is likely to be successful.

Example with larger modulus

Here is an example to recover a message which has been encrypted using RSA to three recipientsusing 512-bit moduli and the common exponent 3 with no random padding. We use our BigDigits library to do the arithmetic. We added a cuberoot function in the latest version 2.3 specifically to solve this typeof problem.

Gauss 10 Crack Cleaner

The example code is in t_bdRsaCrack.c (included in the latest BigDigits distribution).The output of running this code is here. Thanks to Arone Prem Kumar Arokiasami for prompting us to do this.

This shows how easy it is to crack RSA even for realistic key sizes if the sender is careless.

How to prevent this type of attack

Gaussian software, free download

  1. Use a larger exponent, like 65537 (0x10001). This makes it harder to use the above method, but it is much better to...
  2. Add some random bits to the message - at least 64 bits worth. Make sure every message ever encrypted always has different random bytes added. This is known as salting the message and will prevent many otherattacks, too. Obviously, the recipient needs to know how to remove the random bytes after decrypting the message.

Aptech Gauss 10 Crack

For more on weaknesses in RSA and how to combat them, see our RSA algorithm page.

References

  • Menezes, van Oorschot and Vanstone,Handbook of Applied Cryptography,CRC Press LLC, 1997. The complete book is available on-line.
  • M381 Mathematics and Computing: A Third Level Course,Number Theory Handbook,The Open University, 1996.

Contact us

Feedback or questions: Send us a message.

This page first published 23 October 2010 and last updated 5 December 2019

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Gauss 10 Crack Recipe

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