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

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<math>\text{(A)}\ 845 \qquad \text{(B)}\ 945 \qquad \text{(C)}\ 1005 \qquad \text{(D)}\ 1025 \qquad \text{(E)}\ 1045</math>
 
<math>\text{(A)}\ 845 \qquad \text{(B)}\ 945 \qquad \text{(C)}\ 1005 \qquad \text{(D)}\ 1025 \qquad \text{(E)}\ 1045</math>
  
==Solution==
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==Solution 1==
 +
To simplify the problem, we can group 90’s together: <math>90 + 91 + ... + 98 + 99 = 90 \cdot 10 + 1 + 2 + 3 + ... + 8 + 9</math>.
  
===Solution 1===
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<math>90\cdot10=900</math>, and finding <math>1 + 2 + ... + 8 + 9</math> has a trick to it.
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 -> <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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Rearranging the numbers so each pair sums up to 10, we have:
 +
<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>.
  
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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==Solution 2==
 
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We can express each of the terms as a difference from <math>100</math> and then add the negatives using <math>\frac{n(n+1)}{2}=1+2+3+\cdots+(n-1)+n</math> to get the answer.  
945 is <math>\boxed{\text{B}}.</math>
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<cmath>\begin{align*}
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(100-10)+(100-9)+\cdots+(100-1) &= 100\cdot10 -(1+2+\cdots+9+10)\\
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&= 1000 - 55\\
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&= \boxed{\text{(B)}~945}
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\end{align*}</cmath>
  
===Solution 2===
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==Solution 3==
Instead of breaking the sum and then rearranging, we can start by rearranging:
+
Instead of breaking the sum then rearranging, we can rearrange directly:
 
<cmath>\begin{align*}
 
<cmath>\begin{align*}
90+91+92+\cdots +98+99 &= (90+99)+(91+98)+(92+97)+(93+96)+(94+95) \\
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90+91+92+\cdots +98+99 &= (90+99)+(91+98)+(92+97)+(93+96)+(94+95) \\
 
&= 189+189+189+189+189 \\
 
&= 189+189+189+189+189 \\
&= 945\rightarrow \boxed{\text{B}}  
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&= \boxed{\text{(B)}~945}  
 
\end{align*}</cmath>
 
\end{align*}</cmath>
  
===Solution 3===
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==Solution 4==
  
We can use the formula for finite arithmetic sequences.
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The finite arithmetic sequence formula states that the sum in the sequence is equal to <math>\frac{n}{2}\cdot(a_1+a_n)</math> where <math>n</math> is the number of terms in the sequence, <math>a_1</math> is the first term and <math>a_n</math> is the last term.
  
It is <math>\frac{n}{2}\times</math> (<math>a_1+a_n</math>) where <math>n</math> is the number of terms in the sequence, <math>a_1</math> is the first term and <math>a_n</math> is the last term.
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Applying the formula, we have:
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<cmath>\frac{10}{2}\cdot(90+99)=\boxed{\text{(B)}~945}</cmath>
  
Applying it here:
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==Video Solution==
 +
https://youtu.be/1NtsgKc6mXs
  
<math>\frac{10}{2} \times (90+99) = 945 \rightarrow \boxed{B}</math>
+
~savannahsolver
  
 
==See Also==
 
==See Also==

Revision as of 21:54, 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 $\frac{n(n+1)}{2}=1+2+3+\cdots+(n-1)+n$ to get the answer. \begin{align*} (100-10)+(100-9)+\cdots+(100-1) &= 100\cdot10 -(1+2+\cdots+9+10)\\ &= 1000 - 55\\ &= \boxed{\text{(B)}~945} \end{align*}

Solution 3

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

Solution 4

The finite arithmetic sequence formula states that the sum in the sequence is equal to $\frac{n}{2}\cdot(a_1+a_n)$ where $n$ is the number of terms in the sequence, $a_1$ is the first term and $a_n$ is the last term.

Applying the formula, we have: \[\frac{10}{2}\cdot(90+99)=\boxed{\text{(B)}~945}\]

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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