What is the smallest possible positive integer x>5 , that leaves a remainder of 5 whenever it is divided by 14 , 8 or 77 ?

To find the smallest possible positive integer x that satisfies all three conditions (leaving a remainder of 5 when divided by 14, 8, and 77), we can use the concept of the least common multiple (LCM) of these three numbers.

The LCM is the smallest positive integer that is divisible by each of the given numbers. In this case, we need to find the LCM of 14, 8, and 77.

To find the LCM, we can prime factorize each number:

14 = 2 * 7
8 = 2^3
77 = 7 * 11

Now, we need to take the highest power of each prime factor appearing in any of the prime factorizations. Therefore, the LCM will be:

LCM = 2^3 * 7 * 11 = 616

Now, we know that x has a remainder of 5 when divided by 14, 8, and 77. We can express this information as equations:

x ≡ 5 (mod 14)
x ≡ 5 (mod 8)
x ≡ 5 (mod 77)

To solve these congruences, we can use the Chinese Remainder Theorem (CRT), which states that if we have a set of congruences with pairwise relatively prime moduli, there exists a unique solution modulo the product of those moduli.

Since 14, 8, and 77 are pairwise relatively prime, we can use the CRT to find a solution. In this case, the solution can be found by solving the following congruence:

x ≡ 5 (mod 616)

Therefore, the smallest positive integer x that satisfies all three conditions is 5.