1985 AJHSME Problems/Problem 2

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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

Solution 1

One possibility is to simply add them. However, this can be time-consuming, and there are other ways to solve this problem. We find a simpler problem in this problem, and simplify -> $90 + 91 + ... + 98 + 99 = 90 \times 10 + 1 + 2 + 3 + ... + 8 + 9$

We know $90 \times 10$, that's easy: $900$. So how do we find $1 + 2 + ... + 8 + 9$?

We rearrange the numbers to make $(1 + 9) + (2 + 8) + (3 + 7) + (4 + 6) + 5$. You might have noticed that each of the terms we put next to each other add up to 10, which makes for easy adding. $4 \times 10 + 5 = 45$. Adding that on to 900 makes 945.

945 is $\boxed{\text{B}}$

Solution 2

Instead of breaking the sum and then rearranging, we can start by rearranging: \begin{align*} 90+91+92+\cdots +98+99 &=  (90+99)+(91+98)+(92+97)+(93+96)+(94+95) \\ &= 189+189+189+189+189 \\ &= 945\rightarrow \boxed{\text{B}}  \end{align*}

Solution 3

We can use the formula for finite arithmetic sequences.

It is $\frac{n}{2}\times$ ($a_1+a_n$) where $n$ is the number of terms in the sequence, $a_1$ is the first term and $a_n$ is the last term.

Applying it here:

$\frac{10}{2} \times (90+99) = 945 \rightarrow \boxed{B}$

See Also

1985 AJHSME (ProblemsAnswer KeyResources)
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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