AoPSWiki
Our Precalculus course starts on Dec. 4. Master trig, complex numbers, and vectors and matrices in 2 and 3 dimensions. Click here to enroll today!

1985 AJHSME Problems/Problem 2

From AoPSWiki

Contents

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

We could just add them all together. But what would be the point of doing that? So we find a slicker way.

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 \\&= 94...

See Also

1985 AJHSME (ProblemsResources)
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
Visit the AoPS Book Store.
© Copyright 2008 AoPS Incorporated. All Rights Reserved. • FoundationPrivacyContact Us