Difference between revisions of "2016 AMC 8 Problems/Problem 19"
(→Problem) |
(→Problem) |
||
Line 1: | Line 1: | ||
− | |||
− | |||
<math>\textbf{(A)}\mbox{ }360\qquad\textbf{(B)}\mbox{ }388\qquad\textbf{(C)}\mbox{ }412\qquad\textbf{(D)}\mbox{ }416\qquad\textbf{(E)}\mbox{ }424</math> | <math>\textbf{(A)}\mbox{ }360\qquad\textbf{(B)}\mbox{ }388\qquad\textbf{(C)}\mbox{ }412\qquad\textbf{(D)}\mbox{ }416\qquad\textbf{(E)}\mbox{ }424</math> |
Revision as of 17:15, 17 June 2025
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 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 |
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.