Difference between revisions of "1985 AJHSME Problems/Problem 2"

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==Solution 1==
 
==Solution 1==
One possibility is to simply add them. However, this can be time-consuming, and there are other ways to solve this problem.
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To simplify the problem, we can group 90’s together: <math>90 + 91 + ... + 98 + 99 = 90 \cdot 10 + 1 + 2 + 3 + ... + 8 + 9</math>.
We find a simpler problem in this problem, and simplify -> <math>90 + 91 + ... + 98 + 99 = 90 \times 10 + 1 + 2 + 3 + ... + 8 + 9</math>
 
  
We know <math>90 \times 10</math>, that's easy: <math>900</math>. So how do we find <math>1 + 2 + ... + 8 + 9</math>?
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<math>90\cdot10=900</math>, and finding <math>1 + 2 + ... + 8 + 9</math> has a trick to it.
  
We rearrange the numbers to make <math>(1 + 9) + (2 + 8) + (3 + 7) + (4 + 6) + 5</math>. You might have noticed that each of the terms we put next to each other add up to 10, which makes for easy adding. <math>4 \times 10 + 5 = 45</math>. Adding that on to 900 makes 945.
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Rearranging the numbers so each pair sums up to 10, we have:
 
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<cmath>(1 + 9)+(2+8)+(3+7)+(4+6)+5</cmath>. <math>4\cdot10+5 = 45</math>, and <math>900+45=\boxed{\text{B} 945}</math>.
945 is <math>\boxed{\text{B}}</math>
 
  
 
==Solution 2==
 
==Solution 2==

Revision as of 20:28, 1 February 2023

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 $\frac{n(n+1)}{2}$ 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:

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

Video Solution

https://youtu.be/1NtsgKc6mXs

~savannahsolver

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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