Difference between revisions of "1985 AJHSME Problem 2"
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The sum of each pair is <math>189</math> and there are <math>5</math> pairs, so the sum is <math>945</math> and the answer is <math>\boxed{\textbf{(B)}\ 945}</math>. | The sum of each pair is <math>189</math> and there are <math>5</math> pairs, so the sum is <math>945</math> and the answer is <math>\boxed{\textbf{(B)}\ 945}</math>. | ||
| − | ==Cheap | + | ==Solution 3 (Cheap and Quick)== |
We know that <math>10(90) = 900</math> and <math>10(100) = 1000.</math> Quick estimation reveals that this sum is in between these two numbers, so the only answer available is <math>\boxed{\textbf{(B)}\ 945}</math>. | We know that <math>10(90) = 900</math> and <math>10(100) = 1000.</math> Quick estimation reveals that this sum is in between these two numbers, so the only answer available is <math>\boxed{\textbf{(B)}\ 945}</math>. | ||
Revision as of 08:38, 22 March 2025
Contents
[hide]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 
.
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 .
Solution 3 (Cheap and Quick)
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
