What is the largest number that will always divide the sum of any \(19\) consecutive positive integers?
+23246 If the first of 19 consecutive positive integers is x then the 19th integer is x + 18.
Using the formula: Sum = number · (first + last) / 2
Sum = 19( x + (x + 18) ) / 2
Sum = 19( 2x + 18 ) / 2
Sum = 19( 2(x + 9) ) / 2
Sum = 19( x + 9 )
So, I think that the answer is 19.
+36915 Agree with geno.... I read the question as 19 consecutive EVEN integers..... D'Oh! 
