Difference between revisions of "1985 AJHSME Problem 2"
(→Solution 2) |
(→Solution 1) |
||
| Line 8: | Line 8: | ||
We can add as follows: | We can add as follows: | ||
<cmath>90+91+92+93+94+95+96+97+98+99= 10(90) +1+2+3+4+5+6+7+8+9 = 900 + 45 = \boxed{945}</cmath> | <cmath>90+91+92+93+94+95+96+97+98+99= 10(90) +1+2+3+4+5+6+7+8+9 = 900 + 45 = \boxed{945}</cmath> | ||
| − | The answer is <math>\ | + | The answer is <math></math>\boxed{\textbf{(B)}\ 945}$. |
==Solution 2== | ==Solution 2== | ||
Revision as of 08:35, 22 March 2025
Problem
Solution 1
We can add as follows:
![\[90+91+92+93+94+95+96+97+98+99= 10(90) +1+2+3+4+5+6+7+8+9 = 900 + 45 = \boxed{945}\]](http://latex.artofproblemsolving.com/1/d/b/1db4e2342c772b4aed18d27770793629634fa8d5.png)
The answer is $$ (Error compiling LaTeX. Unknown error_msg)\boxed{\textbf{(B)}\ 945}$.
Solution 2
Pair the numbers like so:
![\[(90+99)+(91+98)+(92+97)+(93+96)+(94+95)\]](http://latex.artofproblemsolving.com/9/0/e/90e6ffd152b0189a5f7fd1521ff17296c91172b9.png)
The sum of each pair is 
and there are 
pairs, so the sum is 
and the answer is 
.
Cheap Solution
We know that 
and 
Quick estimation reveals that this sum is in between these two numbers, so the only answer available is 
Solution 3
You see that 
is equal to 
. You can use the formula 
to get 45+900=945
Video Solution
~savannahsolver
