Answer:
37 is the smallest positive integer n such that n(n+1)(n+2) is divisible by 247.
Step-by-step explanation:
First we will find the prime factors of 247:
247 = 13 x 19 (which are both prime).
So now we need to find a number (the smallest one) that is of the form (n)(n+1)(n+2) (the product of three consecutive numbers) and that is divisible by both 13 and 19 (and therefore divisible by 247)
Let's take a look at the multiples of 13: 13, 26, 39, 52...
Let's take a look at the multiples of 19: 19, 38, 57...
We can see that the first time we have two multiples close together are the 38 (for 19) and the 39 (for 17).
So, if our number has both 38 and 39 as factors, then it will be divisible by 247.
However, we need not two but three consecutive numbers, and since we want the number to be the smallest positive integer, we will add 37 (since our other choice would be to add 40 and that would make the number bigger) and thus our number is (37)(38)(39) or in other words (37)(37 + 1)(37 + 2) and therefore this is the smallest positive number such that n(n+1)(n+2) is divisible by 247.
