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

m (→‎Solution 2)
m (→‎Solution 3)
Line 23: Line 23:
  
 
==Solution 3==
 
==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) \\
+
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}}  
+
&= \boxed{\text{(B)}~945}  
 
\end{align*}</cmath>
 
\end{align*}</cmath>
  

Revision as of 20:39, 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

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 (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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