How it Works › Forums › Advanced Problems (AMC 8 Problems 20-25 and beyond) › Unique Properties of Integers Advanced Problems Set 2 › Reply To: Unique Properties of Integers Advanced Problems Set 2
For the first part of the problem, if you look at the formula, it is (n(n+1))/2. So, if n is even, then n+1 is odd and vice versa. If n is even, then the formula is equal to:
(n+1) * n/2. This is a factor of n+1, not n. So, that means n has to be odd to make this work. So, the answer is just all the odd numbers until 100. There are 50 odd numbers.
For the second part of this problem, the most obvious possible answer is 1. 1 (mod 1) = 0. The next possible answer is 5, since n+1 is 6 and 6 divides 6 evenly. Then, the next possible number is 9, where 10*9 (mod 2*3(6)) is 0 because 10 (mod 2) = 0 and 9 (mod 3) equals 0. So, the pattern is 1,5,9,13… until 97. That is 25 numbers.
This response is written by Yash Agarwal.