A6. In how many ways can you make the number 100 by choosing a sequence of consecutive numbers from 1 to 99 and adding them together?
A) 1
B) 2
C) 3
D) 4
E) 5
A6. B) 2

Let the number of numbers we add together be $n$. There are two cases.

$n$ is odd In this case, there is a middle number. Let this number be $k$. The sum of the $n$ numbers is then $n \times k=100$. Since $n$ is an odd divisor of 100 (and greater than 1), $n$ must be 5 or 25. In the first case, we find $k=\frac{100}{5}=20$ and the solution $18+19+20+21+22=100$. In the second case, we find $k=4$, but this does not give a solution because the smallest of the 25 numbers would then be equal to $4-12$, which is a negative number.

$n$ is even Let $n=2m$. The sum of the two middle numbers is an odd number, say $k$. The sum now consists of $m$ pairs of numbers, each with a sum of $k$. This means $100=k \times m$. Since $k$ is an odd divisor of 100 (and greater than 1), it follows that $k=5$ or $k=25$. In the first case, $m=20$ and the two middle numbers are 2 and 3. This does not give a solution, because then the smallest number would be equal to $3-20$, which is negative. In the second case, $m=4$ with the solution $100=9+10+11+12+13+14+15+16$.