To solve the problem, we start by analyzing the remainders when the number (N) is divided by 3, 4, 5, and 6:

Key Observation:

For each divr (d), the remainder (r) satisfies (d - r = 2). This means (N + 2) is divisible by 3, 4, 5, and 6.

We need the smallest positive (N), so (N + 2) must be the least common multiple (LCM) of these divrs.

Calculate LCM of 3, 4, 5, 6:

• Prime factors:
  (3 = 3), (4 = 2^2), (5 = 5), (6 = 2 \times 3).
• LCM = product of highest powers of primes: (2^2 \times 3 \times 5 = 4 \times 3 \times 5 = 60).

Find (N):

(N = LCM - 2 = 60 - 2 = 58).

Answer: 58.

(\boxed{58})

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