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rb276
29.09.2019 •
Mathematics
Solve the congruence x^329 = 452 (mod 1147). [hint. 1147 is not prime.]
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Ответ:
The answer is![x \equiv 763 \:(mod \: 1147)](/tpl/images/0272/9624/6e6f2.png)
Step-by-step explanation:
The following steps will give a solution to the congruence
1. Compute Euler's Phi function
.
We have
by prime factorization, so that ![\phi(1147)=\phi(31)\cdot \phi(37)=30\cdot 36= 1080](/tpl/images/0272/9624/59fc1.png)
because
where p is a prime number.
2. Find positive integers u and v that satisfy
.
We know a solution exists, since
, using the Euclidean algorithm allows us to find the solution ![1 = -151\cdot 329 + 46\cdot 1080](/tpl/images/0272/9624/77552.png)
In order to get positive values for u and v, we modify this solution:
The equation
provides the key to solving the original problem.
3. Compute
by successive squaring. The value obtained gives the solution x.
We have
, so ![x \equiv 452^{929}\:(mod \: 1147)](/tpl/images/0272/9624/66fcc.png)
To use this method start by looking at the exponent 929 and represent it as a sum of powers of 2 this is called the binary expansion of 929. To do this, find the largest power of 2 less than your exponent in this case it’s
. Subtract 512 from 929 getting 417. And continue in this manner to get:
Now
So all you have to do is to calculate the numbers
and multiply them together, then take the product![(mod \: 1147)](/tpl/images/0272/9624/94d77.png)
Ответ:
16
Step-by-step explanation: