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

m
(Added second solution and box)
Line 8: Line 8:
 
==Solution==
 
==Solution==
  
 +
===Solution 1===
 
We could just add them all together. But what would be the point of doing that? So we find a slicker way.
 
We could just add them all together. But what would be the point of doing that? So we find a slicker way.
  
Line 17: Line 18:
  
 
945 is <math>\boxed{\text{B}}</math>
 
945 is <math>\boxed{\text{B}}</math>
 +
 +
===Solution 2===
 +
Instead of breaking the sum and then rearranging, we can start by rearranging:
 +
<cmath>\begin{align*}
 +
90+91+92+\cdots +98+99 &=  (90+99)+(91+98)+(92+97)+(93+96)+(94+95) \\
 +
&= 189+189+189+189+189 \\
 +
&= 945\rightarrow \boxed{\text{A}}
 +
\end{align*}</cmath>
  
 
==See Also==
 
==See Also==
  
[[1985 AJHSME Problems]]
+
{{AJHSME box|year=1985|num-b=1|num-a=3}}
  
 
[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]

Revision as of 08:24, 6 May 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

Solution 1

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 $\boxed{\text{B}}$

Solution 2

Instead of breaking the sum and then rearranging, we can start by rearranging: \begin{align*} 90+91+92+\cdots +98+99 &=  (90+99)+(91+98)+(92+97)+(93+96)+(94+95) \\ &= 189+189+189+189+189 \\ &= 945\rightarrow \boxed{\text{A}}  \end{align*}

See Also

1985 AJHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 1
Followed by
Problem 3
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AJHSME/AMC 8 Problems and Solutions
Invalid username
Login to AoPS