Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus
During AMC testing, the AoPS Wiki is in read-only mode and no edits can be made.

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

(→‎Solution 7 (Brute Force))
Line 7: Line 7:
  
 
==Solution 1==
 
==Solution 1==
−
To simplify the problem, we can group 90’s together: [mathjax]90 + 91 + ... + 98 + 99 = 90 \cdot 10 + 1 + 2 + 3 + ... + 8 + 9[/mathjax].
+
To simplify the problem, we can group <math>90</math>’s together: [mathjax]90 + 91 + ... + 98 + 99 = 90 \cdot 10 + 1 + 2 + 3 + ... + 8 + 9[/mathjax].
  
 
[mathjax]90\cdot10=900[/mathjax], and finding [mathjax]1 + 2 + ... + 8 + 9[/mathjax] has a trick to it.
 
[mathjax]90\cdot10=900[/mathjax], and finding [mathjax]1 + 2 + ... + 8 + 9[/mathjax] has a trick to it.
Line 13: Line 13:
 
Rearranging the numbers so each pair sums up to 10, we have:
 
Rearranging the numbers so each pair sums up to 10, we have:
 
[mathjax display=true](1 + 9)+(2+8)+(3+7)+(4+6)+5[/mathjax]. [mathjax]4\cdot10+5 = 45[/mathjax], and [mathjax]900+45=\boxed{\text{(B)}~945}[/mathjax].
 
[mathjax display=true](1 + 9)+(2+8)+(3+7)+(4+6)+5[/mathjax]. [mathjax]4\cdot10+5 = 45[/mathjax], and [mathjax]900+45=\boxed{\text{(B)}~945}[/mathjax].
−
 
+
~[[shunyipanda]] (Minor edit)
 
==Solution 2==
 
==Solution 2==
 
We can express each of the terms as a difference from <math>100</math> and then add the negatives using <math>\frac{n(n+1)}{2}=1+2+3+\cdots+(n-1)+n</math> to get the answer.  
 
We can express each of the terms as a difference from <math>100</math> and then add the negatives using <math>\frac{n(n+1)}{2}=1+2+3+\cdots+(n-1)+n</math> to get the answer.  
Line 41: Line 41:
  
 
~ [https://artofproblemsolving.com/wiki/index.php/User:Cxsmi cxsmi]
 
~ [https://artofproblemsolving.com/wiki/index.php/User:Cxsmi cxsmi]
 +
~[[shunyipanda]] (Minor edit)
  
 
== Solution 6 (Estimate) ==
 
== Solution 6 (Estimate) ==
Line 46: Line 47:
 
Notice the sum is larger than <math>90 \times 10</math> and less than <math>100 \times 10</math>
 
Notice the sum is larger than <math>90 \times 10</math> and less than <math>100 \times 10</math>
  
−
The only choice that works is 945, thus the answer is <math>\boxed{\textbf{(B)}\ 945}</math>
+
The only choice that works is <math>945</math>, thus the answer is <math>\boxed{\textbf{(B)}\ 945}</math>
  
 
~ lovelearning999
 
~ lovelearning999
Line 52: Line 53:
 
== Solution 7 (Brute Force) ==
 
== Solution 7 (Brute Force) ==
  
−
Add all the numbers to get 945, or <math>\boxed{\textbf{(D)}\ 945}</math>
+
Add all the numbers to get <math>945</math>, or <math>\boxed{\textbf{(B)}\ 945}</math>
  
−
~ lovelearning999
+
~ lovelearning999 ~shunyipanda (Minor edit)
  
 +
== Solution 8 (Simpler Solution 1) ==
 +
Many people know that <math>1+2+3...+9</math> is <math>45</math>, and if we seperate this arithmetic sequence from the rest of the <math>90</math>'s, we get <math>45+900=\boxed{\textbf{(B)}\ 945}</math>.
 +
~[[shunyipanda]]
 
==Video Solution by BoundlessBrain!==
 
==Video Solution by BoundlessBrain!==
 
https://youtu.be/8bVNfa-yEoM
 
https://youtu.be/8bVNfa-yEoM

Revision as of 19:40, 22 October 2025

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 1

To simplify the problem, we can group $90$’s together: [mathjax]90 + 91 + ... + 98 + 99 = 90 \cdot 10 + 1 + 2 + 3 + ... + 8 + 9[/mathjax].

[mathjax]90\cdot10=900[/mathjax], and finding [mathjax]1 + 2 + ... + 8 + 9[/mathjax] has a trick to it.

Rearranging the numbers so each pair sums up to 10, we have: [mathjax display=true](1 + 9)+(2+8)+(3+7)+(4+6)+5[/mathjax]. [mathjax]4\cdot10+5 = 45[/mathjax], and [mathjax]900+45=\boxed{\text{(B)}~945}[/mathjax]. ~shunyipanda (Minor edit)

Solution 2

We can express each of the terms as a difference from $100$ and then add the negatives using $\frac{n(n+1)}{2}=1+2+3+\cdots+(n-1)+n$ to get the answer. \begin{align*} (100-10)+(100-9)+\cdots+(100-1) &= 100\cdot10 -(1+2+\cdots+9+10)\\ &= 1000 - 55\\ &= \boxed{\text{(B)}~945} \end{align*}

Solution 3

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

Solution 4

The finite arithmetic sequence formula states that the sum in the sequence is equal to $\frac{n}{2}\cdot(a_1+a_n)$ where $n$ is the number of terms in the sequence, $a_1$ is the first term and $a_n$ is the last term.

Applying the formula, we have: \[\frac{10}{2}\cdot(90+99)=\boxed{\text{(B)}~945}\]

Solution 5

The expression is equal to the sum of integers from $1$ to $99$ minus the sum of integers from $1$ to $89$, so it is equal to $\frac{99(100)}{2} - \frac{89(90)}{2} = 4950 - 4005 = \boxed{\text{(B)}~945}$.

~ cxsmi ~shunyipanda (Minor edit)

Solution 6 (Estimate)

Notice the sum is larger than $90 \times 10$ and less than $100 \times 10$

The only choice that works is $945$, thus the answer is $\boxed{\textbf{(B)}\ 945}$

~ lovelearning999

Solution 7 (Brute Force)

Add all the numbers to get $945$, or $\boxed{\textbf{(B)}\ 945}$

~ lovelearning999 ~shunyipanda (Minor edit)

Solution 8 (Simpler Solution 1)

Many people know that $1+2+3...+9$ is $45$, and if we seperate this arithmetic sequence from the rest of the $90$'s, we get $45+900=\boxed{\textbf{(B)}\ 945}$. ~shunyipanda

Video Solution by BoundlessBrain!

https://youtu.be/8bVNfa-yEoM

Video Solution

https://youtu.be/1NtsgKc6mXs

~savannahsolver

See Also

1985 AJHSME (Problems • Answer Key • Resources)
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


These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1985_AJHSME_Problems/Problem_2&oldid=263894"