How do you solve congruences?

To solve a linear congruence ax ≡ b (mod N), you can multiply by the inverse of a if gcd(a,N) = 1; otherwise, more care is needed, and there will either be no solutions or several (exactly gcd(a,N) total) solutions for x mod N.

How do you solve Chinese remainder theorem problems?

Example: Solve the simultaneous congruences x ≡ 6 (mod 11), x ≡ 13 (mod 16), x ≡ 9 (mod 21), x ≡ 19 (mod 25). Solution: Since 11, 16, 21, and 25 are pairwise relatively prime, the Chinese Remainder Theorem tells us that there is a unique solution modulo m, where m = 11⋅16⋅21⋅25 = 92400.

How do you solve a Mod?

How to calculate the modulo – an example

1. Start by choosing the initial number (before performing the modulo operation).
2. Choose the divisor.
3. Divide one number by the other, rounding down: 250 / 24 = 10 .
4. Multiply the divisor by the quotient.
5. Subtract this number from your initial number (dividend).

What are the last two digits of 49 19 using Chinese remainder theorem?

Step-by-step explanation: The Chinese remainder theorem provides with a unique solution to simultaneous linear congruences with the coprime modulo. The modulo generally being 100. Hence, the last two digits of 49^19 is 49.

Why Chinese remainder theorem is used?

The Chinese remainder theorem is widely used for computing with large integers, as it allows replacing a computation for which one knows a bound on the size of the result by several similar computations on small integers.

What does mod 9 mean?

Modular 9 arithmetic is the arithmetic of the remainders after division by 9. For example, the remainder for 12 after division by 9 is 3.

Is there a solution to solving simultaneous congruences?

Although Bill Cook’s answer is completely, 100% correct (and based on the proof of the Chinese Remainder Theorem), one can also work with the congruences successively; we know from the CRT that a solution exists. Starting from x ≡ 3 ( mod 7), this means that x is of the form x = 7 k + 3 for some integer k.

How to solve the second congruence of modular arithmetic?

Substituting into the second congruence, we get 7 k + 3 ≡ 2 ( mod 5). Since 7 ≡ 2 ( mod 5), and 3 ≡ − 2 ( mod 5), this is equivalent to 2 k − 2 ≡ 2 ( mod 5). Dividing through by 2 (which we can do since gcd ( 2, 5) = 1) we get k − 1 ≡ 1 ( mod 5), or k ≡ 2 ( mod 5). That is, k is of the form k = 5 r + 2 for some integer r.

How to solve the problem of linear congruence?

Trying to figure out how to solve linear congruence by following through the sample solution to the following problem: I understand up to this point, but the next line I don’t get: Where is the × 3, × 2, × 1 from? Is it just because there are 3 terms, so it starts from 3 then 2 then 1? And where is the 52 coming from?

Which is the form of the second congruence?

Starting from x ≡ 3 ( mod 7), this means that x is of the form x = 7 k + 3 for some integer k. Substituting into the second congruence, we get