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Difference between revisions of "1985 AJHSME Problems/Problem 2"
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Line 16: Line 16:
 
945 is <math>\boxed{\text{B}}</math>
 
945 is <math>\boxed{\text{B}}</math>
  
−
==Solution 2==
+
==Solution 4==
 +
We can express each of the terms as a difference from 100 and then add the negatives using <math>\frac{n(n+1)}{2}</math> to get the answer.
 +
<cmath>\begin{align*}
 +
(100-10)+(100-9)+\cdots + (100-1) &= 100 \times 10 -(1+2+\cdots +9+10)\\
 +
&= 1000 - 55\\
 +
&= 945 \rightarrow \boxed{\text{B}}
 +
\end{align*}</cmath>
 +
 
 +
==Solution 3==
 
Instead of breaking the sum and then rearranging, we can start by rearranging:
 
Instead of breaking the sum and then rearranging, we can start by rearranging:
 
<cmath>\begin{align*}
 
<cmath>\begin{align*}
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\end{align*}</cmath>
 
\end{align*}</cmath>
  
−
==Solution 3==
+
==Solution 4==
  
 
We can use the formula for finite arithmetic sequences.
 
We can use the formula for finite arithmetic sequences.

Revision as of 01:18, 10 June 2021

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

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 4

We can express each of the terms as a difference from 100 and then add the negatives using $\frac{n(n+1)}{2}$ to get the answer. \begin{align*} (100-10)+(100-9)+\cdots + (100-1) &= 100 \times 10 -(1+2+\cdots +9+10)\\ &= 1000 - 55\\ &= 945 \rightarrow \boxed{\text{B}} \end{align*}

Solution 3

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 4

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 (Problems • Answer Key • Resources)
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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