# 1985 AJHSME Problems/Problem 2

## Problem

$90+91+92+93+94+95+96+97+98+99=$

$\text{(A)}\ 845 \qquad \text{(B)}\ 945 \qquad \text{(C)}\ 1005 \qquad \text{(D)}\ 1025 \qquad \text{(E)}\ 1045$

## Solution 1

To simplify the problem, we can group 90’s together: [mathjax]90 + 91 + ... + 98 + 99 = 90 \cdot 10 + 1 + 2 + 3 + ... + 8 + 9[/mathjax].

[mathjax]90\cdot10=900[/mathjax], and finding [mathjax]1 + 2 + ... + 8 + 9[/mathjax] has a trick to it.

Rearranging the numbers so each pair sums up to 10, we have: [mathjax display=true](1 + 9)+(2+8)+(3+7)+(4+6)+5[/mathjax]. [mathjax]4\cdot10+5 = 45[/mathjax], and [mathjax]900+45=\boxed{\text{(B)}~945}[/mathjax].

## Solution 2

We can express each of the terms as a difference from $100$ and then add the negatives using $\frac{n(n+1)}{2}=1+2+3+\cdots+(n-1)+n$ to get the answer. \begin{align*} (100-10)+(100-9)+\cdots+(100-1) &= 100\cdot10 -(1+2+\cdots+9+10)\\ &= 1000 - 55\\ &= \boxed{\text{(B)}~945} \end{align*}

## Solution 3

Instead of breaking the sum then rearranging, we can rearrange directly: \begin{align*} 90+91+92+\cdots +98+99 &= (90+99)+(91+98)+(92+97)+(93+96)+(94+95) \\ &= 189+189+189+189+189 \\ &= \boxed{\text{(B)}~945} \end{align*}

## Solution 4

The finite arithmetic sequence formula states that the sum in the sequence is equal to $\frac{n}{2}\cdot(a_1+a_n)$ where $n$ is the number of terms in the sequence, $a_1$ is the first term and $a_n$ is the last term.

Applying the formula, we have: $$\frac{10}{2}\cdot(90+99)=\boxed{\text{(B)}~945}$$

## Solution 5

The expression is equal to the sum of integers from $1$ to $99$ minus the sum of integers from $1$ to $89$, so it is equal to $\frac{99(100)}{2} - \frac{89(90)}{2} = 4950 - 4005 = \boxed{\text{(B)}~945}$.

~savannahsolver