Difference between revisions of "1985 AJHSME Problems/Problem 2"

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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)
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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 -> <math>90 + 91 + ... + 98 + 99 = 90 \times 10 + 1 + 2 + 3 + ... + 8 + 9</math>
 +
 
 +
We know <math>90 \times 10</math>, that's easy - <math>900</math>. So how do we find <math>1 + 2 + ... + 8 + 9</math>?
 +
 
 +
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 we put next to each other add up to 10, which makes for easy adding. <math>4 \times 10 + 5 = 45</math>. Adding that on to 900 makes 945.
 +
 
 +
945 is <math>\boxed{\text{B}}</math>
  
 
==See Also==
 
==See Also==
  
 
[[1985 AJHSME Problems]]
 
[[1985 AJHSME Problems]]

Revision as of 21:12, 12 January 2009

Problem

$90+91+92+93+94+95+96+97+98+99=$


$\text{(A)}\ 845 \qquad \text{(B)}\ 945 \qquad \text{(C)}\ 1005 \qquad \text{(D)}\ 1025 \qquad \text{(E)}\ 1045$

Solution

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 -> $90 + 91 + ... + 98 + 99 = 90 \times 10 + 1 + 2 + 3 + ... + 8 + 9$

We know $90 \times 10$, that's easy - $900$. So how do we find $1 + 2 + ... + 8 + 9$?

We rearrange the numbers to make $(1 + 9) + (2 + 8) + (3 + 7) + (4 + 6) + 5$. You might have noticed that each of the terms we put next to each other add up to 10, which makes for easy adding. $4 \times 10 + 5 = 45$. Adding that on to 900 makes 945.

945 is $\boxed{\text{B}}$

See Also

1985 AJHSME Problems

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