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

Solution 1

We could just add them all together. But what would be the point of doing that? So we find a slicker way.

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: $\begin{align*} 90+91+92+\cdots +98+99 &= (90+99)+(91+98)+(92+97)+(93+96)+(94+95) \\ &= 189+189+189+189+189 \\ &= 945\rightarrow \boxed{\text{A}} \end{align*}$

1985 AJHSME (Problems • Answer Key • Resources)
Preceded by Problem 1	Followed by Problem 3
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1985 AJHSME Problems/Problem 2

Contents

Problem

Solution

Solution 1

Solution 2

See Also