If a, b, c are three distinct natural numbers, all less than 100, such that $$\mid a - b \mid + \mid b - c \mid = \mid c - a \mid$$, then the maximum possible value of b is ______

$$\mid a - b \mid + \mid b - c \mid = \mid c - a \mid$$. This equation is only valid when b lies between a and c. Thus, there are two cases possible - $$a<b<c$$ or $$c<b<a$$

Case - 1 :- $$a<b<c$$

Now, we need to find the maximum value of b. All, a, b, and c are natural numbers is less than 100, and we $$c>b$$. Thus, if we assume the value of c = 99 and b = 98 (which are the two maximum possible values), then -

$$\mid a-98\mid+\mid98-99\mid=\mid99-a\mid$$, and we assumed that a < b, thus a < 98.

=> $$98-a+1=99-a$$ => 0 = 0

Thus, any value of a < 98 is possible for this equation.

Therefore, c = 99, b = 98 and a < 98

Case - 2 :- $$c<b<a$$

Now, we need to find the maximum value of b. All, a, b, and c are natural numbers is less than 100, and we $$a>b$$. Thus, if we assume the value of a = 99 and b = 98 (which are the two maximum possible values), then -

$$\mid99-98\mid+\mid98-c\mid=\mid c-99\mid$$, and we assumed that c < b, thus c < 98.

=>$$1+98-c=99-c$$ => 0 = 0

Thus, any value of c < 98 is possible for this equation.

Therefore, a = 99, b = 98 and c < 98

Thus, in both cases, the maximum value of b = 98.