1985 AJHSME 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

We can add as follows: $90+91+92+93+94+95+96+97+98+99= 10(90) +1+2+3+4+5+6+7+8+9 = 900 + 45 = \boxed{945}$ The answer is $\boxed{\textbf{(B)}\ 945}$ .

Solution 2

Pair the numbers like so: $(90+99)+(91+98)+(92+97)+(93+96)+(94+95)$ The sum of each pair is $189$ and there are $5$ pairs, so the sum is $945$ and the answer is $\boxed{\textbf{(B)}\ 945}$ .

Solution 3 (Cheap and Quick)

We know that $10(90) = 900$ and $10(100) = 1000.$ Quick estimation reveals that this sum is in between these two numbers, so the only answer available is $\boxed{\textbf{(B)}\ 945}$ .

Solution $3\frac{1}{2}$

You see that $90+91+92+93+94+95+96+97+98+99$ is equal to $0+1+3+4+5+6+7+8+9+900$ . You can use the formula $\frac{n(n+1)}{2}$ to get $45+900=\boxed{\textbf{(B)}\ 945}$ .

Video Solution

https://youtu.be/1NtsgKc6mXs

~savannahsolver

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