Difference between revisions of "1985 AJHSME Problems/Problem 2"
5849206328x (talk | contribs) m |
5849206328x (talk | contribs) (Added second solution and box) |
||
Line 8: | Line 8: | ||
==Solution== | ==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 could just add them all together. But what would be the point of doing that? So we find a slicker way. | ||
Line 17: | Line 18: | ||
945 is <math>\boxed{\text{B}}</math> | 945 is <math>\boxed{\text{B}}</math> | ||
+ | |||
+ | ===Solution 2=== | ||
+ | Instead of breaking the sum and then rearranging, we can start by rearranging: | ||
+ | <cmath>\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{A}} | ||
+ | \end{align*}</cmath> | ||
==See Also== | ==See Also== | ||
− | + | {{AJHSME box|year=1985|num-b=1|num-a=3}} | |
[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] |
Revision as of 07:24, 6 May 2009
Problem
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 ->
We know , that's easy - . So how do we find ?
We rearrange the numbers to make . You might have noticed that each of the terms we put next to each other add up to 10, which makes for easy adding. . Adding that on to 900 makes 945.
945 is
Solution 2
Instead of breaking the sum and then rearranging, we can start by rearranging:
See Also
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 |