Difference between revisions of "2016 AMC 8 Problems/Problem 19"
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~MrThinker | ~MrThinker | ||
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+ | ==Solution 3== | ||
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+ | Let <math>x</math> be the smallest number. The equation will become, <math>x+(x+2)+(x+4)+\cdots +(x+48)=10000</math>. After you combine like terms, you get <math>25x+(50*12)=10000</math> which turns into <math>10000-600=25x</math>. <math>25x=9400</math>, so <math>x=376</math>. Then you add <math>376+48 = \boxed{\textbf{(E)}\ 424}</math>. | ||
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+ | ~AfterglowBlaziken | ||
==Video Solution== | ==Video Solution== |
Revision as of 18:37, 22 December 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
Solution 3
Let be the smallest number. The equation will become,
. After you combine like terms, you get
which turns into
.
, so
. Then you add
.
~AfterglowBlaziken
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.