Difference between revisions of "2016 AMC 8 Problems/Problem 19"
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Let <math>x</math> be the largest number. Then, <math>x+(x-2)+(x-4)+\cdots +(x-50)=10000</math>. Factoring this gives | Let <math>x</math> be the largest number. Then, <math>x+(x-2)+(x-4)+\cdots +(x-50)=10000</math>. Factoring this gives | ||
<math>2\left(\frac{x}{2} + \left(\frac{x}{2} - 1\right) + \left(\frac{x}{2} - 2\right) +\cdots + \left(\frac{x}{2} - 25\right)\right)=1000</math>. Grouping like terms gives <math>25\left(\frac{x}{2}\right) - 300=5000</math>, and continuing down the line, we find <math>x=\boxed{\textbf{(E)}\ 424}</math>. | <math>2\left(\frac{x}{2} + \left(\frac{x}{2} - 1\right) + \left(\frac{x}{2} - 2\right) +\cdots + \left(\frac{x}{2} - 25\right)\right)=1000</math>. Grouping like terms gives <math>25\left(\frac{x}{2}\right) - 300=5000</math>, and continuing down the line, we find <math>x=\boxed{\textbf{(E)}\ 424}</math>. | ||
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| + | ~MrThinker | ||
==Video Solution== | ==Video Solution== | ||
Revision as of 10:20, 1 November 2022
Problem
The sum of 
consecutive even integers is 
. What is the largest of these consecutive integers?
Solution 1
Let 
be the 13th consecutive even integer that's being added up. By doing this, we can see that the sum of all 25 even numbers will simplify to 
since 
. Now, 
Remembering that this is the 13th integer, we wish to find the 25th, which is 
.
Solution 2
Let 
be the largest number. Then, 
. Factoring this gives
. Grouping like terms gives 
, and continuing down the line, we find 
.
~MrThinker
Video Solution
~savannahsolver
See Also
| 2016 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 18 |
Followed by Problem 20 | |
| 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. 
