Final answer:
The smallest positive integer N that satisfies the given divisibility conditions by the squares of the primes 2, 3, 5, and 7 is 1764. It is the only option where N, N+1, and N+2 are each divisible by a different one of these squares.
Step-by-step explanation:
The question requires finding the smallest positive integer N such that N, N+1, and N+2 are each divisible by the squares of different primes: 2, 3, 5, and 7. To solve this problem, we need to consider the divisibility conditions for each number:
- A number is divisible by 2^2 (4) if its last two digits form a number divisible by 4.
- A number is divisible by 3^2 (9) if the sum of its digits is divisible by 9.
- A number is divisible by 5^2 (25) if its last two digits are 00, 25, 50, or 75.
- A number is divisible by 7^2 (49) only if it satisfies the divisibility rule of 49, which is more complex and typically requires direct division to verify.
Through testing each option, we determine that:
- 630 is divisible by 5^2 and 7^2, but 631 and 632 are not divisible by 2^2 or 3^2.
- 882 is divisible by 2^2, 883 is not divisible by 3^2 or 5^2, and 884 is divisible by 2^2 again, failing the requirement.
- 1260 is divisible by 2^2, 1261 is not divisible by 3^2, and 1262 is divisible by 2^2, so it also fails.
- 1764 is divisible by 7^2, 1765 is divisible by 5^2, and 1766 is divisible by 2^2, meeting all the requirements.
Therefore, the smallest positive integer N with the given property is 1764.