Difference between revisions of "1985 AJHSME Problem 2"
Coolmath34 (talk | contribs) (Created page with "==Problem== <math>90+91+92+93+94+95+96+97+98+99=</math> <math>\text{(A)}\ 845 \qquad \text{(B)}\ 945 \qquad \text{(C)}\ 1005 \qquad \text{(D)}\ 1025 \qquad \text{(E)}\ 1045<...") |
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==Cheap Solution== | ==Cheap Solution== | ||
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>\text{(B)}.</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>\text{(B)}.</math> | ||
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| + | ==Video Solution== | ||
| + | https://youtu.be/1NtsgKc6mXs | ||
| + | |||
| + | ~savannahsolver | ||
Latest revision as of 07:36, 7 January 2023
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
Cheap Solution
We know that 
and 
Quick estimation reveals that this sum is in between these two numbers, so the only answer available is
Video Solution
~savannahsolver
