Difference between revisions of "1985 AJHSME Problems/Problem 2"
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==Solution 2== | ==Solution 2== | ||
| − | We can express each of the terms as a difference from 100 and then add the negatives using <math>\frac{n(n+1)}{2}</math> to get the answer. | + | 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*} | <cmath>\begin{align*} | ||
| − | (100-10)+(100-9)+\cdots + (100-1) &= 100 \ | + | (100-10)+(100-9)+\cdots+(100-1) &= 100\cdot10 -(1+2+\cdots+9+10)\\ |
&= 1000 - 55\\ | &= 1000 - 55\\ | ||
| − | &= | + | &= \boxed{\text{(B)}~945} |
\end{align*}</cmath> | \end{align*}</cmath> | ||
Revision as of 20:37, 1 February 2023
Problem
Solution 1
To simplify the problem, we can group 90’s together: 
.
, and finding 
has a trick to it.
Rearranging the numbers so each pair sums up to 10, we have:
![\[(1 + 9)+(2+8)+(3+7)+(4+6)+5\]](http://latex.artofproblemsolving.com/1/5/1/151425ffc3700df99ccddfbfcf3c4452458051d3.png)
. 
, and 
.
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 and then rearranging, we can start by rearranging:
Solution 4
We can use the formula for finite arithmetic sequences.
It is 
(
) where 
is the number of terms in the sequence, 
is the first term and 
is the last term.
Applying it here:
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. 
