Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2015 AMC 8 Problems/Problem 14"

(→‎Solution 2)
(→‎Solution 2)
Line 8: Line 8:
 
===Solution 2===
 
===Solution 2===
 
If the four consecutive odd integers are <math>2n-3,~ 2n-1, ~2n+1</math> and <math>2n+3</math> then the sum is <math>8n</math>. All the integers are divisible by <math>8</math> except <math>100</math>, so the answer is <math>\boxed{\textbf{(D)}~100}</math>.
 
If the four consecutive odd integers are <math>2n-3,~ 2n-1, ~2n+1</math> and <math>2n+3</math> then the sum is <math>8n</math>. All the integers are divisible by <math>8</math> except <math>100</math>, so the answer is <math>\boxed{\textbf{(D)}~100}</math>.
 +
 +
 +
 +
===Solution 3===
 +
If the four consecutive odd integers are <math>a,~ a+2, ~a+4</math> and <math>a+6</math> and the sum is <math>b</math> then <math>b</math> divided by <math>4</math> is equal to <math>a+3</math>. This means that <math>a+3</math> must be even. The only integer that does not give an even integer when divided by <math>4</math> is <math>100</math>, so the answer is <math>\boxed{\textbf{(D)}~100}</math>.
  
 
==See Also==
 
==See Also==

Revision as of 13:58, 29 November 2015

Which of the following integers cannot be written as the sum of four consecutive odd integers?

$\textbf{(A)}\text{ 16}\quad\textbf{(B)}\text{ 40}\quad\textbf{(C)}\text{ 72}\quad\textbf{(D)}\text{ 100}\quad\textbf{(E)}\text{ 200}$

Solution 1

Let our $4$ numbers be $n, n+2, n+4, n+6$, where $n$ is odd. Then our sum is $4n+12$. The only answer choice that cannot be written as $4n+12$, where $n$ is odd, is $\boxed{\textbf{(D)}\text{ 100}}$.

Solution 2

If the four consecutive odd integers are $2n-3,~ 2n-1, ~2n+1$ and $2n+3$ then the sum is $8n$. All the integers are divisible by $8$ except $100$, so the answer is $\boxed{\textbf{(D)}~100}$.


Solution 3

If the four consecutive odd integers are $a,~ a+2, ~a+4$ and $a+6$ and the sum is $b$ then $b$ divided by $4$ is equal to $a+3$. This means that $a+3$ must be even. The only integer that does not give an even integer when divided by $4$ is $100$, so the answer is $\boxed{\textbf{(D)}~100}$.

See Also

2015 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 13 		Followed by
Problem 15
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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_8_Problems/Problem_14&oldid=73313"