Difference between revisions of "1985 AJHSME Problems/Problem 2"
5849206328x (talk | contribs) (New page: ==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</math> ==So...) |
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==Solution== | ==Solution== | ||
− | + | We could just add them all together. But what would the point be of doing that? So we find a slicker way.<br><br>We find a simpler problem in this problem, and simplify -> <math>90 + 91 + ... + 98 + 99 = 90 \times 10 + 1 + 2 + 3 + ... + 8 + 9</math><br>We know <math>90 \times 10</math>, that's easy - <math>900</math>. So how do we find <math>1 + 2 + ... + 8 + 9</math>?<br><br>Well once again (not once again if you didn't read the solution for #1), commutative property comes to the rescue. We rearrange the numbers to make <math>(1 + 9) + (2 + 8) + (3 + 7) + (4 + 6) + 5</math>. You might have noticed that each of the terms i put next to each other add up to 10, which make for easy adding. <math>4 \times 10 + 5 = 45</math>. Adding that on to 900 makes 945.<br><br>945 is (B) | |
==See Also== | ==See Also== | ||
[[1985 AJHSME Problems]] | [[1985 AJHSME Problems]] |
Revision as of 21:03, 12 January 2009
Problem
Solution
We could just add them all together. But what would the point be 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 ?
Well once again (not once again if you didn't read the solution for #1), commutative property comes to the rescue. We rearrange the numbers to make . You might have noticed that each of the terms i put next to each other add up to 10, which make for easy adding. . Adding that on to 900 makes 945.
945 is (B)