Solution:

Given, the remainder when the positive integer m is divided by 7 is x, and the remainder when m is divided by 14 is x+7.

Let's use the Chinese Remainder Theorem to solve this problem.

According to the Chinese Remainder Theorem, if we have a set of congruences of the form:

x ≡ a1 (mod m1)
x ≡ a2 (mod m2)
⋮
x ≡ ak (mod mk)

where m1, m2, …, mk are pairwise coprime integers (i.e., they have no common factors other than 1), then there exists a unique solution x modulo M = m1m2⋯mk.

We can apply this theorem to our problem as follows:

Let a1 = x, m1 = 7 (since the remainder when m is divided by 7 is x)
Let a2 = x+7, m2 = 14 (since the remainder when m is divided by 14 is x+7)

Note that m1 and m2 are not coprime, since they have a common factor of 7. However, we can still use the Chinese Remainder Theorem by dividing both sides of the second congruence by 7, which gives:

x ≡ (x+7) (mod 14)
x ≡ x+7 (mod 14)
0 ≡ 7 (mod 14)

This last congruence is true for all integers x, since 7 is a multiple of 14. Therefore, we can ignore it and focus on the first two congruences:

x ≡ a1 (mod m1)
x ≡ a2 (mod m2)

Substituting the values we have for a1, m1, a2, and m2, we get:

x ≡ x (mod 7)
x ≡ x+7 (mod 14)

To find the unique solution modulo M = m1m2 = 7×14 = 98, we can use the following formula:

x ≡ a1M2y1 + a2M1y2 (mod M)

where M1 = M/m1 = 14, M2 = M/m2 = 7, y1 and y2 are integers such that M1y1 ≡ 1 (mod m1) and M2y2 ≡ 1 (mod m2).

To find y1 and y2, we can use the Extended Euclidean Algorithm:

14y1 ≡ 1 (mod 7)
y1 ≡ 1 (mod 7)
y1 = 1

7y2 ≡ 1 (mod 14)
7y2 ≡ 15 (mod 14) (since 15 is also a solution)
y2 ≡ 2 (mod 14/7)
y2 ≡ 2 (mod 2)
y2 = 2

Substituting these values into the formula, we get:

x ≡ x×982 + (x+7)×1472 (mod 98)
x ≡ 53 (mod 98)

Therefore, the possible values for m are those that leave a remainder of 53 when divided by 98. Option B, 53, is the only one that satisfies this condition