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I just need a solution but it seems like I cannot get. I tried Euclid algorithm and Euler's theorem but no progress.

Arturo Magidin
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Mire
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    See the linked dupes for most all known methods, e.g. by inverse reciprocity or twiddling $\bmod 45!:\ x\equiv \dfrac{34}7\equiv\dfrac{-11}7\equiv \dfrac{-56}7\equiv -8\ \ $ – Bill Dubuque May 15 '23 at 06:23

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To find a solution for the congruence equation $7X\equiv 34 \pmod{45}$, we can use modular arithmetic.

First, we need to find the modular multiplicative inverse of 7 modulo 45. The modular inverse of 7 exists if and only if 7 and 45 are coprime (i.e., their greatest common divisor is 1).

Checking their gcd, we have:

$\text{gcd}(7, 45) = 1$

Since the gcd is 1, 7 and 45 are coprime. Thus, the modular inverse of 7 modulo 45 exists.

Now, we can find the modular inverse using the extended Euclidean algorithm or by inspection. In this case, the modular inverse of 7 modulo 45 is 13.

To solve the congruence equation, we multiply both sides by the modular inverse:

$7X \cdot 13 \equiv 34 \cdot 13 \pmod{45}$

$91X \equiv 442 \pmod{45}$

Reducing the coefficients:

$1X \equiv 37 \pmod{45}$

So, the solution to the congruence equation $7X\equiv 34 \pmod{45}$ is $X \equiv 37 \pmod{45}$.

MatrixMagician
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