Number Systems

Step 1: Count the number of 1-digit numbers.
There are 9 one-digit numbers: . Step 2: Count the number of 2-digit numbers.
For a 2-digit number, the first digit can be any of (9 choices), and the second digit can be any of the remaining 9 digits (0-9, except the first digit). Thus, there are: Step 3: Count the number of 3-digit numbers.
For a 3-digit number less than 500, the first digit can be any of (4 choices), the second digit can be any of the remaining 9 digits (0-9 except the first digit), and the third digit can be any of the remaining 8 digits. Thus, there are: Step 4: Add the total count.
The total number of natural numbers less than 500 with no repeated digits is: Thus, the number of natural numbers less than 500 in which no two digits are repeated is .

We need to determine the maximum value of k for which n(n²+3) is divisible by k for all n ∈ ℕ.

Step 1: Consider the expression n(n²+3). Expanding this gives us:

Step 2: Analyze divisibility for a few small values of n.

• If n = 1, then 1³ + 3×1 = 4.
• If n = 2, then 2³ + 3×2 = 14.
• If n = 3, then 3³ + 3×3 = 36.

Step 3: Determine the greatest common divisor (GCD) of these values.

Observing these results, the expressions consistently give a GCD of 2. Thus, n(n²+3) is divisible by 2 for any natural number n, but not necessarily by any larger number for all n.

Conclusion: The maximum value of k such that n(n²+3) is divisible by k for all n ∈ ℕ is 2.