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: $90 + 91 + ... + 98 + 99 = 90 \cdot 10 + 1 + 2 + 3 + ... + 8 + 9$ .

$90\cdot10=900$ , and finding $1 + 2 + ... + 8 + 9$ has a trick to it.

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

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 $\frac{n}{2}\cdot(a_1+a_n)3 where$ 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: $\frac{10}{2}\cdot(90+99)=\boxed{\text{(B)}~945}$ .

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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