To solve the system of congruences using Brahmagupta’s procedure (which is closely related to the method of successive substitution or the Chinese remainder theorem), we are given the following:

$N \equiv 23 \pmod{137}$ $N \equiv 0 \pmod{60}$

We need to find $N$ that satisfies both congruences.

Step 1: Express $N$ in terms of one congruence

From the second congruence $N \equiv 0 \pmod{60}$, we can express $N$ as: $N = 60k \quad \text{for some integer } k.$

Step 2: Substitute into the other congruence

Substitute $N = 60k$ into the first congruence $N \equiv 23 \pmod{137}$: $60k \equiv 23 \pmod{137}.$

Step 3: Solve the linear congruence

Now we solve the congruence $60k \equiv 23 \pmod{137}$. This is a linear congruence, and we need to find $k$.

First, find the inverse of 60 modulo 137. We use the extended Euclidean algorithm to compute the greatest common divisor (gcd) of 60 and 137 and express it in the form: $60x + 137y = \gcd(60, 137).$ Since the gcd is 1 (as 60 and 137 are coprime), we can find $x$ which will be the inverse of 60 modulo 137.

Step 3.1: Extended Euclidean Algorithm

We apply the Euclidean algorithm to 60 and 137:

$137 = 2 \times 60 + 17$ $60 = 3 \times 17 + 9$ $17 = 1 \times 9 + 8$ $9 = 1 \times 8 + 1$ $8 = 8 \times 1 + 0.$ Thus, the gcd is 1.

Now, we backtrack to express 1 as a linear combination of 60 and 137:

$1 = 9 - 1 \times 8$ $1 = 9 - 1 \times (17 - 1 \times 9) = 2 \times 9 - 1 \times 17$ $1 = 2 \times (60 - 3 \times 17) - 1 \times 17 = 2 \times 60 - 7 \times 17$ $1 = 2 \times 60 - 7 \times (137 - 2 \times 60) = 16 \times 60 - 7 \times 137.$

Thus, $16 \times 60 - 7 \times 137 = 1$, so the inverse of 60 modulo 137 is 16.

Step 3.2: Solve for $k$

Now that we have the inverse, we solve for $k$: $k \equiv 16 \times 23 \pmod{137}.$ Calculate $16 \times 23 = 368$. Now reduce modulo 137: $368 \div 137 \approx 2.686 \quad \text{so} \quad 368 - 2 \times 137 = 368 - 274 = 94.$ Thus, $k \equiv 94 \pmod{137}$.

Step 4: Find $N$

Now that we have $k \equiv 94 \pmod{137}$, we substitute this back into $N = 60k$: $N = 60 \times 94 = 5640.$

Thus, the solution to the system of congruences is $N = 5640$.

Step 5: Verify the solution

• Check $N \equiv 23 \pmod{137}$: $5640 \div 137 \approx 41.168 \quad \text{so} \quad 5640 - 41 \times 137 = 5640 - 5617 = 23,$ which is correct.

• Check $N \equiv 0 \pmod{60}$: $5640 \div 60 = 94,$ which is an integer, so $N \equiv 0 \pmod{60}$ is also correct.

Thus, the solution is $N = 5640$.

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Would you like a detailed explanation on any part of the process?

Here are 5 related questions:

1. How can we use the Chinese Remainder Theorem for systems with more congruences?
2. What is the significance of the extended Euclidean algorithm in solving linear congruences?
3. How do we know when two numbers are coprime, and why is this important for modular inverses?
4. How would the solution change if the modulus 137 were not coprime with 60?
5. How can Brahmagupta's procedure be generalized to other number systems?

Tip: When solving modular systems, always ensure the moduli are coprime for simpler computations!