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"
(→‎Video Solution)
 
(16 intermediate revisions by 9 users not shown)
Line 1: Line 1:
 +
==Problem==
 +
 
Which of the following integers cannot be written as the sum of four consecutive odd integers?
 
Which of the following integers cannot be written as the sum of four consecutive odd integers?
  
 
<math>\textbf{(A)}\text{ 16}\quad\textbf{(B)}\text{ 40}\quad\textbf{(C)}\text{ 72}\quad\textbf{(D)}\text{ 100}\quad\textbf{(E)}\text{ 200}</math>
 
<math>\textbf{(A)}\text{ 16}\quad\textbf{(B)}\text{ 40}\quad\textbf{(C)}\text{ 72}\quad\textbf{(D)}\text{ 100}\quad\textbf{(E)}\text{ 200}</math>
  
−
===Solution===
+
==Solutions==
−
Let our <math>4</math> numbers be <math>n, n+2, n+4, n+6</math>, where <math>n</math> is odd. Then our sum is <math>4n+12</math>. The only answer choice that cannot be written as <math>4n+12</math>, where <math>n</math> is odd, is <math>\boxed{\textbf{(D)}\text{ 100}}</math>.
+
 
 +
===Solution 1===
 +
Let our <math>4</math> numbers be <math>n, n+2, n+4, n+6</math>, where <math>n</math> is odd. Then, our sum is <math>4n+12</math>. The only answer choice that cannot be written as <math>4n+12</math>, where <math>n</math> is odd, is <math>\boxed{\textbf{(D)}\text{ 100}}</math>.
 +
 
 +
===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>\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>, the sum is <math>4a+12</math>, and <math>4a+12</math> divided by <math>4</math> gives <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>.
 +
 
 +
===Solution 4===
 +
From Solution 1, we have the sum of the <math>4</math> numbers to be equal to <math>4n + 12</math>. Taking mod 8 gives us <math>4n + 4 \equiv b \pmod8</math> for some residue <math>b</math> and for some odd integer <math>n</math>. Since <math>n \equiv 1 \pmod{2}</math>, we can express it as the equation <math>n = 2a + 1</math> for some integer <math>a</math>. Multiplying 4 to each side of the equation yields <math>4n = 8a + 4</math>, and taking mod 8 gets us <math>4n \equiv 4 \pmod{8}</math>, so <math>b = 0</math>. All the answer choices except choice D is a multiple of 8, and since 100 satisfies all the conditions the answer is <math>\boxed{\textbf{(D)}~100}</math>.
 +
 
 +
===Solution 5===
 +
Since they want CONSECUTIVE odd numbers, it won't be hard to just list the sums out:
 +
<cmath>16=1+3+5+7</cmath>
 +
<cmath>40=7+9+11+13</cmath>
 +
<cmath>72=15+17+19+21</cmath>
 +
<cmath>200=47+49+51+53</cmath>
 +
All of the answer choices can be a sum of consecutive odd numbers except <math>100</math>, so the answer is <math>\boxed{\textbf{(D)} 100}</math>.
 +
 
 +
==Video Solution (HOW TO THINK CREATIVELY!!!)==
 +
https://youtu.be/Xc7EYnJL5_c
 +
 
 +
~Education, the Study of Everything
 +
 
 +
==Video Solution==
 +
https://youtu.be/khCUYxEJFQg
 +
 
 +
~savannahsolver
 +
 
 
==See Also==
 
==See Also==
  
 
{{AMC8 box|year=2015|num-b=13|num-a=15}}
 
{{AMC8 box|year=2015|num-b=13|num-a=15}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 15:46, 14 May 2023

Problem

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

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

Solutions

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 $\boxed{\textbf{(D)}~100}$.

Solution 3

If the four consecutive odd integers are $a,~ a+2, ~a+4$ and $a+6$, the sum is $4a+12$, and $4a+12$ divided by $4$ gives $a+3$. 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 $\boxed{\textbf{(D)}~100}$.

Solution 4

From Solution 1, we have the sum of the $4$ numbers to be equal to $4n + 12$. Taking mod 8 gives us $4n + 4 \equiv b \pmod8$ for some residue $b$ and for some odd integer $n$. Since $n \equiv 1 \pmod{2}$, we can express it as the equation $n = 2a + 1$ for some integer $a$. Multiplying 4 to each side of the equation yields $4n = 8a + 4$, and taking mod 8 gets us $4n \equiv 4 \pmod{8}$, so $b = 0$. All the answer choices except choice D is a multiple of 8, and since 100 satisfies all the conditions the answer is $\boxed{\textbf{(D)}~100}$.

Solution 5

Since they want CONSECUTIVE odd numbers, it won't be hard to just list the sums out: \[16=1+3+5+7\] \[40=7+9+11+13\] \[72=15+17+19+21\] \[200=47+49+51+53\] All of the answer choices can be a sum of consecutive odd numbers except $100$, so the answer is $\boxed{\textbf{(D)} 100}$.

Video Solution (HOW TO THINK CREATIVELY!!!)

https://youtu.be/Xc7EYnJL5_c

~Education, the Study of Everything

Video Solution

https://youtu.be/khCUYxEJFQg

~savannahsolver

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=193126"