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

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<math>\text{(A)}\ 845 \qquad \text{(B)}\ 945 \qquad \text{(C)}\ 1005 \qquad \text{(D)}\ 1025 \qquad \text{(E)}\ 1045</math>
 
<math>\text{(A)}\ 845 \qquad \text{(B)}\ 945 \qquad \text{(C)}\ 1005 \qquad \text{(D)}\ 1025 \qquad \text{(E)}\ 1045</math>
  
−
==Solution==
+
==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].
  
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===Solution 1===
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[mathjax]90\cdot10=900[/mathjax], and finding [mathjax]1 + 2 + ... + 8 + 9[/mathjax] has a trick to it.
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One possibility is to simply add them. However, this can be time-consuming, and there are other ways to solve this problem.
 
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We find a simpler problem in this problem, and simplify -> <math>90 + 91 + ... + 98 + 99 = 90 \times 10 + 1 + 2 + 3 + ... + 8 + 9</math>
 
  
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We know <math>90 \times 10</math>, that's easy - <math>900</math>. So how do we find <math>1 + 2 + ... + 8 + 9</math>?
+
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].
  
−
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.
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==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.  
−
945 is <math>\boxed{\text{B}}.</math>
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<cmath>\begin{align*}
 +
(100-10)+(100-9)+\cdots+(100-1) &= 100\cdot10 -(1+2+\cdots+9+10)\\
 +
&= 1000 - 55\\
 +
&= \boxed{\text{(B)}~945}
 +
\end{align*}</cmath>
  
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===Solution 2===
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==Solution 3==
−
Instead of breaking the sum and then rearranging, we can start by rearranging:
+
Instead of breaking the sum then rearranging, we can rearrange directly:
 
<cmath>\begin{align*}
 
<cmath>\begin{align*}
−
90+91+92+\cdots +98+99 &= (90+99)+(91+98)+(92+97)+(93+96)+(94+95) \\
+
90+91+92+\cdots +98+99 &= (90+99)+(91+98)+(92+97)+(93+96)+(94+95) \\
 
&= 189+189+189+189+189 \\
 
&= 189+189+189+189+189 \\
−
&= 945\rightarrow \boxed{\text{B}}  
+
&= \boxed{\text{(B)}~945}  
 
\end{align*}</cmath>
 
\end{align*}</cmath>
  
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===Solution 3===
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==Solution 4==
 +
 
 +
The finite arithmetic sequence formula states that the sum in the sequence is equal to <math>\frac{n}{2}\cdot(a_1+a_n)</math> where <math>n</math> is the number of terms in the sequence, <math>a_1</math> is the first term and <math>a_n</math> is the last term.
 +
 
 +
Applying the formula, we have:
 +
<cmath>\frac{10}{2}\cdot(90+99)=\boxed{\text{(B)}~945}</cmath>
 +
 
 +
==Solution 5==
 +
The expression is equal to the sum of integers from <math>1</math> to <math>99</math> minus the sum of integers from <math>1</math> to <math>89</math>, so it is equal to <math>\frac{99(100)}{2} - \frac{89(90)}{2} = 4950 - 4005 = \boxed{\text{(B)}~945}</math>.
 +
 
 +
~ [https://artofproblemsolving.com/wiki/index.php/User:Cxsmi cxsmi]
 +
 
 +
== Solution 6 (Estimate) ==
 +
 
 +
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>
 +
 
 +
~ lovelearning999
 +
 
 +
== Solution 7 (Brute Force) ==
 +
 
 +
Add all the numbers to get 945, or <math>\boxed{\textbf{(D)}\ 945}</math>
  
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We can use the formula for finite arithmetic sequences.
+
~ lovelearning999
  
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It is <math>\frac{n}{2}\times</math> (First Term+Last Term) where <math>n</math> is the number of terms in the sequence.  
+
==Video Solution by BoundlessBrain!==
 +
https://youtu.be/8bVNfa-yEoM
  
−
Applying it here:
+
==Video Solution==
 +
https://youtu.be/1NtsgKc6mXs
  
−
<math>\frac{10}{2} \times (90+99) = 945 \rightarrow \boxed{B}</math>
+
~savannahsolver
  
 
==See Also==
 
==See Also==

Latest revision as of 21:04, 2 October 2024

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].

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

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 $\boxed{\textbf{(D)}\ 945}$

~ lovelearning999

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


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