Two friends A and B start running around a circular track at the same time,…

Concept: When two runners move in the SAME direction around a circular track with distinct integer speeds p and q, they meet repeatedly forever, but only at a limited number of DISTINCT points before the pattern repeats. That count of distinct meeting points equals |p − q| divided by gcd(p, q) — the gap between the speeds, reduced by whatever they share in common.

Application: Here p = 6 and q = b, so the number of distinct meeting points is |b − 6| / gcd(6, b). We need this to equal 2, i.e. |b − 6| = 2·gcd(6, b). Since gcd(6, b) must be one of 6's own divisors — 1, 2, 3, or 6 — check each case:

1. gcd(6, b) = 1 → |b − 6| = 2 → b = 4 or 8. But gcd(6, 4) = 2 and gcd(6, 8) = 2, neither equals the assumed 1, so both fail the consistency check.

2. gcd(6, b) = 2 → |b − 6| = 4 → b = 2 or 10. Checking: gcd(6, 2) = 2 and gcd(6, 10) = 2 — both are consistent, so both are valid.

3. gcd(6, b) = 3 → |b − 6| = 6 → b = 0 or 12. b = 0 is not a natural number, and gcd(6, 12) = 6, not 3, so this case gives no valid b.

4. gcd(6, b) = 6 → |b − 6| = 12 → b = −6 or 18. b = −6 is invalid, but gcd(6, 18) = 6, so b = 18 is valid.

So the only values satisfying the equation are b = 2, 10, 18 — three natural numbers under 30.

Cross-check: Take b = 2 on a unit-length track. The relative speed is 4, so they meet every 1/4 s. At meeting number n, A has covered 6·(n/4) = 1.5n of the track, i.e. position (1.5n mod 1). For n = 0, 1, 2, 3 this gives positions 0, 0.5, 0, 0.5 — exactly 2 distinct points, confirming the formula. By contrast, at the boundary value b = 4 (from the rejected gcd = 1 case), the relative speed is 2 and every meeting lands back at position 0 — only 1 distinct point — which is why that branch was correctly discarded above.

Hence there are 3 possible values of b (namely 2, 10, and 18).