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

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

@@ Line 6: / Line 6: @@
 <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].
-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)
+[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 <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.
+<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>
+==Solution 3==
+Instead of breaking the sum then rearranging, we can rearrange directly:
+<cmath>\begin{align*}
++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*}</cmath>
+==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>
+~ lovelearning999
+==Video Solution by BoundlessBrain!==
+https://youtu.be/8bVNfa-yEoM
+==Video Solution==
+https://youtu.be/1NtsgKc6mXs
+~savannahsolver
 ==See Also==
-[[1985 AJHSME Problems]]
+{{AJHSME box|year=1985|num-b=1|num-a=3}}
+{{MAA Notice}}
+[[Category:Introductory Algebra Problems]]

1985 AJHSME (Problems • Answer Key • Resources)
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