Solution:

To solve this question, we need to use the Chinese Remainder Theorem. According to the theorem, if we have a system of linear congruences:

x ≡ a1 (mod m1)
x ≡ a2 (mod m2)
...
x ≡ an (mod mn)

where m1, m2, ..., mn are pairwise coprime, then the system has a unique solution modulo M = m1 × m2 × ... × mn.

In this case, we have the following system of linear congruences:

x ≡ 5 (mod 70)
x ≡ 8 (mod 125)

We need to find the largest number which divides 70 and 125 and satisfies the above system of linear congruences.

Step 1: Finding the Greatest Common Divisor (GCD)

We will use the Euclidean algorithm to find the GCD of 70 and 125.

125 = 1 × 70 + 55
70 = 1 × 55 + 15
55 = 3 × 15 + 10
15 = 1 × 10 + 5
10 = 2 × 5 + 0

Therefore, the GCD of 70 and 125 is 5.

Step 2: Finding the Solution

Since the GCD of 70 and 125 is 5, we can write:

70 = 5 × 14
125 = 5 × 25

We can now rewrite the system of linear congruences as:

x ≡ 5 (mod 5 × 14)
x ≡ 8 (mod 5 × 25)

Simplifying, we get:

x ≡ 5 (mod 14)
x ≡ 8 (mod 25)

Now, we can use the Chinese Remainder Theorem to find the solution:

Let M = 14 × 25 = 350.

We need to find the inverse of 14 modulo 25 and the inverse of 25 modulo 14.

Inverse of 14 modulo 25:

25 = 1 × 14 + 11
14 = 1 × 11 + 3
11 = 3 × 3 + 2
3 = 1 × 2 + 1
2 = 2 × 1 + 0

From the above calculation, we get:

1 = 3 − 1 × 2
1 = 3 − 1 × (11 − 3 × 3)
1 = 4 × 3 − 1 × 11
1 = 4 × (25 − 1 × 14) − 1 × 11
1 = 4 × 25 − 5 × 14

Therefore, the inverse of 14 modulo 25 is 4.

Inverse of 25 modulo 14:

14 = 1 × 14 + 0
25 = 1 × 14 + 11
14 = 1 × 11 + 3
11 = 3 × 3 + 2
3 = 1 × 2 + 1
2 = 2 × 1 + 0

From the above calculation, we get:

1 = 3 − 1 × 2
1 = 3 − 1 × (11 − 3 × 3)
1 = 4 × 3 −