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

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==Solution 1==
 
==Solution 1==
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To simplify the problem, we can group 90’s together: <math>90 + 91 + ... + 98 + 99 = 90 \cdot 10 + 1 + 2 + 3 + ... + 8 + 9</math>.
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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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<math>90\cdot10=900</math>, and finding <math>1 + 2 + ... + 8 + 9</math> has a trick to it.
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[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:
 
Rearranging the numbers so each pair sums up to 10, we have:
−
<cmath>(1 + 9)+(2+8)+(3+7)+(4+6)+5</cmath>. <math>4\cdot10+5 = 45</math>, and <math>900+45=\boxed{\text{(B)}~945}</math>.
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[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==
 
==Solution 2==
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Applying the formula, we have:
 
Applying the formula, we have:
 
<cmath>\frac{10}{2}\cdot(90+99)=\boxed{\text{(B)}~945}</cmath>
 
<cmath>\frac{10}{2}\cdot(90+99)=\boxed{\text{(B)}~945}</cmath>
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==Solution 5==
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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>.
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~ [https://artofproblemsolving.com/wiki/index.php/User:Cxsmi cxsmi]
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== Solution 6 (Estimate) ==
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Notice the sum is larger than <math>90 \times 10</math> and less than <math>100 \times 10</math>
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The only choice that works is 945, thus the answer is <math>\boxed{\textbf{(B)}\ 945}</math>
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~ lovelearning999
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== Solution 7 (Brute Force) ==
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Add all the numbers to get 945, or <math>\boxed{\textbf{(D)}\ 945}</math>
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~ lovelearning999
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==Video Solution by BoundlessBrain!==
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https://youtu.be/8bVNfa-yEoM
  
 
==Video Solution==
 
==Video Solution==

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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