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

Solution 1

One possibility is to simply add them. However, this can be time-consuming, and there are other ways to solve this problem. We find a simpler problem in this problem, and simplify -> $90 + 91 + ... + 98 + 99 = 90 \times 10 + 1 + 2 + 3 + ... + 8 + 9$

We know $90 \times 10$ , that's easy - $900$ . So how do we find $1 + 2 + ... + 8 + 9$ ?

We rearrange the numbers to make $(1 + 9) + (2 + 8) + (3 + 7) + (4 + 6) + 5$ . You might have noticed that each of the terms we put next to each other add up to 10, which makes for easy adding. $4 \times 10 + 5 = 45$ . Adding that on to 900 makes 945.

945 is $\boxed{\text{B}}.$

Solution 2

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

Solution 3

We can use the formula for finite arithmetic sequences.

It is $\frac{n}{2}\times$ ( $a_1+a_n$ ) where $n$ is the number of terms in the sequence, $a_1$ is the first term and $a_n$ is the last term.

Applying it here:

$\frac{10}{2} \times (90+99) = 945 \rightarrow \boxed{B}$

1985 AJHSME (Problems • Answer Key • Resources)
Preceded by Problem 1	Followed by Problem 3
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AJHSME/AMC 8 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

Page

Toolbox

Search