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

m (Solution 4)
m (Solution 4)
(One intermediate revision by the same user not shown)
Line 32: Line 32:
 
==Solution 4==
 
==Solution 4==
  
The finite arithmetic sequence formula states that the sum in the sequence is equal to <math>\frac{n}{2}\cdot(a_1+a_n)3 where </math>n<math> is the number of terms in the sequence, </math>a_1<math> is the first term and </math>a_n$ is the last term.
+
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.
  
 
Applying the formula, we have:
 
Applying the formula, we have:
<cmath>\frac{10}{2}\cdot(90+99)=\boxed{\text{(B)}~945</cmath>.
+
<cmath>\frac{10}{2}\cdot(90+99)=\boxed{\text{(B)}~945}</cmath>
  
 
==Video Solution==
 
==Video Solution==

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


The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png