Problem: Find the largest positive integer that will divide 398, 436 and 542 leaving reminder 7 , 11 and 15 respectively.

Solution:
To find the largest positive integer that will divide 398, 436 and 542 leaving reminders 7, 11 and 15 respectively, we can use the Chinese Remainder Theorem.

Step 1: Express the given system of congruences as a system of linear equations. We have:

398 = a (mod x)
436 = b (mod x)
542 = c (mod x)

where a = 7, b = 11, and c = 15.

Step 2: Solve the system of linear equations using the method of substitution. We can solve for a in the first equation to get:

a = 398 - kx

Substituting this expression for a into the second equation, we get:

436 = (398 - kx) (mod x)

Expanding this expression and simplifying, we get:

kx = 40 (mod x)

Similarly, substituting the expression for a into the third equation, we get:

542 = (398 - kx) (mod x)

Expanding and simplifying, we get:

kx = 144 (mod x)

Step 3: Solve for x using the Chinese Remainder Theorem. We have:

x | (40 - 144) and x | gcd(40, 144)

Simplifying, we get:

x | 104 and x | 8

The largest positive integer that divides all three numbers leaving the given remainders is the greatest common divisor (GCD) of these numbers. Therefore, we have:

GCD(398 - 7, 436 - 11, 542 - 15) = GCD(391, 425, 527)

We can use the Euclidean algorithm to find the GCD:

GCD(391, 425) = GCD(391, 34)
GCD(391, 34) = GCD(11, 34)
GCD(11, 34) = GCD(11, 12)
GCD(11, 12) = GCD(1, 12)
GCD(1, 12) = 1

Therefore, the largest positive integer that will divide 398, 436 and 542 leaving reminders 7, 11 and 15 respectively is 1.