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== | + | ==Solution 1== |
+ | To simplify the problem, we can group 90’s together: [mathjax]90 + 91 + ... + 98 + 99 = 90 \cdot 10 + 1 + 2 + 3 + ... + 8 + 9[/mathjax]. | ||
− | = | + | [mathjax]90\cdot10=900[/mathjax], and finding [mathjax]1 + 2 + ... + 8 + 9[/mathjax] has a trick to it. |
− | |||
− | + | Rearranging the numbers so each pair sums up to 10, we have: | |
+ | [mathjax display=true](1 + 9)+(2+8)+(3+7)+(4+6)+5[/mathjax]. [mathjax]4\cdot10+5 = 45[/mathjax], and [mathjax]900+45=\boxed{\text{(B)}~945}[/mathjax]. | ||
− | We | + | ==Solution 2== |
− | + | 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. | |
− | + | <cmath>\begin{align*} | |
+ | (100-10)+(100-9)+\cdots+(100-1) &= 100\cdot10 -(1+2+\cdots+9+10)\\ | ||
+ | &= 1000 - 55\\ | ||
+ | &= \boxed{\text{(B)}~945} | ||
+ | \end{align*}</cmath> | ||
− | + | ==Solution 3== | |
− | + | Instead of breaking the sum then rearranging, we can rearrange directly: | |
− | |||
− | Instead of breaking the sum | ||
<cmath>\begin{align*} | <cmath>\begin{align*} | ||
− | 90+91+92+\cdots +98+99 &= | + | 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 \\ | ||
− | &= | + | &= \boxed{\text{(B)}~945} |
\end{align*}</cmath> | \end{align*}</cmath> | ||
+ | |||
+ | ==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)</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: | ||
+ | <cmath>\frac{10}{2}\cdot(90+99)=\boxed{\text{(B)}~945}</cmath> | ||
+ | |||
+ | ==Solution 5== | ||
+ | The expression is equal to the sum of integers from <math>1</math> to <math>99</math> minus the sum of integers from <math>1</math> to <math>89</math>, so it is equal to <math>\frac{99(100)}{2} - \frac{89(90)}{2} = 4950 - 4005 = \boxed{\text{(B)}~945}</math>. | ||
+ | |||
+ | ~ [https://artofproblemsolving.com/wiki/index.php/User:Cxsmi cxsmi] | ||
+ | |||
+ | ==Video Solution by BoundlessBrain!== | ||
+ | https://youtu.be/8bVNfa-yEoM | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/1NtsgKc6mXs | ||
+ | |||
+ | ~savannahsolver | ||
==See Also== | ==See Also== | ||
Line 31: | Line 54: | ||
{{AJHSME box|year=1985|num-b=1|num-a=3}} | {{AJHSME box|year=1985|num-b=1|num-a=3}} | ||
+ | |||
+ | {{MAA Notice}} | ||
[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] |
Latest revision as of 13:43, 4 April 2024
Contents
Problem
Solution 1
To simplify the problem, we can group 90’s together: [mathjax]90 + 91 + ... + 98 + 99 = 90 \cdot 10 + 1 + 2 + 3 + ... + 8 + 9[/mathjax].
[mathjax]90\cdot10=900[/mathjax], and finding [mathjax]1 + 2 + ... + 8 + 9[/mathjax] has a trick to it.
Rearranging the numbers so each pair sums up to 10, we have: [mathjax display=true](1 + 9)+(2+8)+(3+7)+(4+6)+5[/mathjax]. [mathjax]4\cdot10+5 = 45[/mathjax], and [mathjax]900+45=\boxed{\text{(B)}~945}[/mathjax].
Solution 2
We can express each of the terms as a difference from and then add the negatives using
to get the answer.
Solution 3
Instead of breaking the sum then rearranging, we can rearrange directly:
Solution 4
The finite arithmetic sequence formula states that the sum in the sequence is equal to where
is the number of terms in the sequence,
is the first term and
is the last term.
Applying the formula, we have:
Solution 5
The expression is equal to the sum of integers from to
minus the sum of integers from
to
, so it is equal to
.
~ cxsmi
Video Solution by BoundlessBrain!
Video Solution
~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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.