1985 AJHSME Problems/Problem 2

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 edits)

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 edits)

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 $45+900=\boxed{\textbf{(B)}\ 945}$ .

Solution 9 (Sigma)

We can write this as $\sum_{k=90}^{99}(k)$ . Solve the sigma to get $\boxed{\textbf{(B)}\ 945}$ . ~shunyipanda

~shunyipanda

Video Solution by BoundlessBrain!

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

Video Solution

https://youtu.be/1NtsgKc6mXs

~savannahsolver

1985 AJHSME (Problems • Answer Key • Resources)
Preceded by Problem 1	Followed by Problem 3
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All AJHSME/AMC 8 Problems and Solutions

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