Hi, regarding the following question:
If a is an odd integer, then the product of a, a-1 and a+1 is always divisible by at least how many integers from 1 to 9?
The answer in the solution is 6, however, what about the case where a = 1?
Let’s check for each integer 1 to 9 by either showing why the product is always divisible by that number or providing a counterexample:
1: Yes, every integer is divisible by 1, so the product of 3 integers will be
2: Yes, both a-1 and a+1 will be even since a is odd, so there will be a factor of 2 in the prime factorization.
3: Yes, a-1, a, a+1 are 3 consecutive integers. Thus, one of them must be a multiple of 3, which will produce a 3 in the prime factorization.
4: Yes, both a-1 and a+1 will be even for the same reason discussed in 2. Thus, there will be two factors of 2, which will produce a 4 in the prime factorization.
5: No, (2)(3)(4) = 24 → not divisible by 5
6: Yes, since 2 and 3 are factors, then 6 will be.
7: No, (2)(3)(4) = 24 → not divisible by 7
8: Yes, because either a-1 or a+1 will be a multiple of 4 and the other will just be a multiple of 2, producing an 8 in the prime factorization.
Here are examples: (2)(3)(4), (4)(5)(6), (6)(7)(8).
9: No, (2)(3)(4) = 24 → not divisible by 9
Thus, we have found 6 numbers: 1,2,3,4,6,8.
When a = 1, we have 1,0,2 as the numbers. Their product is 0, which is divisible by every number. However, we are looking for which numbers the product is always divisible by. Thus, we have seen cases where the product is divisible by less than 9 numbers. This case thus does not tell us any new information.