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 18"
(→‎See Also)
 
(13 intermediate revisions by 9 users not shown)
Line 1: Line 1:
 
==Problem==
 
==Problem==
−
An arithmetic sequence is a sequence in which each term after the first is obtained by adding a constant to the previous term. Each row and each column in this <math>5\times5</math> array is an arithmetic sequence with five terms. The square in the center is labelled <math>X</math> as shown. What is the value of <math>X</math>?
+
An arithmetic sequence is a sequence in which each term after the first is obtained by adding a constant to the previous term. For example, <math>2,5,8,11,14</math> is an arithmetic sequence with five terms, in which the first term is <math>2</math> and the constant added is <math>3</math>. Each row and each column in this <math>5\times5</math> array is an arithmetic sequence with five terms. The square in the center is labelled <math>X</math> as shown. What is the value of <math>X</math>?
  
 
<math>\textbf{(A) }21\qquad\textbf{(B) }31\qquad\textbf{(C) }36\qquad\textbf{(D) }40\qquad \textbf{(E) }42</math>
 
<math>\textbf{(A) }21\qquad\textbf{(B) }31\qquad\textbf{(C) }36\qquad\textbf{(D) }40\qquad \textbf{(E) }42</math>
Line 54: Line 54:
 
</asy>
 
</asy>
  
−
==Solution 1==
+
==Solutions==
 +
 
 +
===Solution 1===
 
We begin filling in the table.  The top row has a first term <math>1</math> and a fifth term <math>25</math>, so we have the common difference is <math>\frac{25-1}4=6</math>.  This means we can fill in the first row of the table:
 
We begin filling in the table.  The top row has a first term <math>1</math> and a fifth term <math>25</math>, so we have the common difference is <math>\frac{25-1}4=6</math>.  This means we can fill in the first row of the table:
 
<asy>
 
<asy>
Line 166: Line 168:
 
We must find the third term of the arithmetic sequence with a first term of <math>13</math> and a fifth term of <math>49</math>.  The common difference of this sequence is <math>\frac{49-13}4=9</math>, so  the third term is <math>13+2\cdot 9=\boxed{\textbf{(B) }31}</math>.
 
We must find the third term of the arithmetic sequence with a first term of <math>13</math> and a fifth term of <math>49</math>.  The common difference of this sequence is <math>\frac{49-13}4=9</math>, so  the third term is <math>13+2\cdot 9=\boxed{\textbf{(B) }31}</math>.
  
−
==Solution 2==
+
===Solution 2===
 
The middle term of the first row is <math>\frac{25+1}{2}=13</math>, since the middle number is just the average in an arithmetic sequence. Similarly, the middle of the bottom row is <math>\frac{17+81}{2}=49</math>. Applying this again for the middle column, the answer is <math>\frac{49+13}{2}=\boxed{\textbf{(B)}~31}</math>.
 
The middle term of the first row is <math>\frac{25+1}{2}=13</math>, since the middle number is just the average in an arithmetic sequence. Similarly, the middle of the bottom row is <math>\frac{17+81}{2}=49</math>. Applying this again for the middle column, the answer is <math>\frac{49+13}{2}=\boxed{\textbf{(B)}~31}</math>.
  
−
==Solution 3==
+
===Solution 3===
  
 
The value of <math>X</math> is simply the average of the average values of both diagonals that contain <math>X</math>. This is <math>\frac{\frac{1+81}{2}+\frac{17+25}{2}}{2} =\frac{\frac{82}{2}+\frac{42}{2}}{2} = \frac{41+21}{2} = \boxed{\textbf{(B)}~31}</math>
 
The value of <math>X</math> is simply the average of the average values of both diagonals that contain <math>X</math>. This is <math>\frac{\frac{1+81}{2}+\frac{17+25}{2}}{2} =\frac{\frac{82}{2}+\frac{42}{2}}{2} = \frac{41+21}{2} = \boxed{\textbf{(B)}~31}</math>
  
−
==See Also==
+
==Video Solution (HOW TO THINK CRITICALLY!!!)==
 +
https://youtu.be/1BUbtuK_MXI
 +
 
 +
~Education, the Study of Everything
 +
 
 +
 
 +
== Video Solution==
 +
https://youtu.be/tKsYSBdeVuw?t=1258
  
 +
== Video Solution ==
 +
https://youtu.be/QFC4VmFa9Kc
  
 +
~savannahsolver
 +
== See Also==
 
{{AMC8 box|year=2015|num-b=17|num-a=19}}
 
{{AMC8 box|year=2015|num-b=17|num-a=19}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 18:56, 25 September 2024

Problem

An arithmetic sequence is a sequence in which each term after the first is obtained by adding a constant to the previous term. For example, $2,5,8,11,14$ is an arithmetic sequence with five terms, in which the first term is $2$ and the constant added is $3$. Each row and each column in this $5\times5$ array is an arithmetic sequence with five terms. The square in the center is labelled $X$ as shown. What is the value of $X$?

$\textbf{(A) }21\qquad\textbf{(B) }31\qquad\textbf{(C) }36\qquad\textbf{(D) }40\qquad \textbf{(E) }42$

[asy] size(3.85cm); label("$X$",(2.5,2.1),N); for (int i=0; i<=5; ++i) draw((i,0)--(i,5), linewidth(.5)); for (int j=0; j<=5; ++j) draw((0,j)--(5,j), linewidth(.5)); void draw_num(pair ll_corner, int num) { label(string(num), ll_corner + (0.5, 0.5), p = fontsize(19pt)); } draw_num((0,0), 17); draw_num((4, 0), 81); draw_num((0, 4), 1); draw_num((4,4), 25); void foo(int x, int y, string n) { label(n, (x+0.5,y+0.5), p = fontsize(19pt)); } foo(2, 4, " "); foo(3, 4, " "); foo(0, 3, " "); foo(2, 3, " "); foo(1, 2, " "); foo(3, 2, " "); foo(1, 1, " "); foo(2, 1, " "); foo(3, 1, " "); foo(4, 1, " "); foo(2, 0, " "); foo(3, 0, " "); foo(0, 1, " "); foo(0, 2, " "); foo(1, 0, " "); foo(1, 3, " "); foo(1, 4, " "); foo(3, 3, " "); foo(4, 2, " "); foo(4, 3, " "); [/asy]

Solutions

Solution 1

We begin filling in the table. The top row has a first term $1$ and a fifth term $25$, so we have the common difference is $\frac{25-1}4=6$. This means we can fill in the first row of the table: [asy] size(3.85cm); label("$X$",(2.5,2.1),N); for (int i=0; i<=5; ++i) draw((i,0)--(i,5), linewidth(.5)); for (int j=0; j<=5; ++j) draw((0,j)--(5,j), linewidth(.5)); void draw_num(pair ll_corner, int num) { label(string(num), ll_corner + (0.5, 0.5), p = fontsize(19pt)); } draw_num((0,0), 17); draw_num((4, 0), 81); draw_num((1,4), 7); draw_num((2,4), 13); draw_num((3,4), 19); draw_num((0, 4), 1); draw_num((4,4), 25); void foo(int x, int y, string n) { label(n, (x+0.5,y+0.5), p = fontsize(19pt)); } foo(2, 4, " "); foo(3, 4, " "); foo(0, 3, " "); foo(2, 3, " "); foo(1, 2, " "); foo(3, 2, " "); foo(1, 1, " "); foo(2, 1, " "); foo(3, 1, " "); foo(4, 1, " "); foo(2, 0, " "); foo(3, 0, " "); foo(0, 1, " "); foo(0, 2, " "); foo(1, 0, " "); foo(1, 3, " "); foo(1, 4, " "); foo(3, 3, " "); foo(4, 2, " "); foo(4, 3, " "); [/asy]

The fifth row has a first term of $17$ and a fifth term of $81$, so the common difference is $\frac{81-17}4=16$. We can fill in the fifth row of the table as shown: [asy] size(3.85cm); label("$X$",(2.5,2.1),N); for (int i=0; i<=5; ++i) draw((i,0)--(i,5), linewidth(.5)); for (int j=0; j<=5; ++j) draw((0,j)--(5,j), linewidth(.5)); void draw_num(pair ll_corner, int num) { label(string(num), ll_corner + (0.5, 0.5), p = fontsize(19pt)); } draw_num((0,0), 17); draw_num((4, 0), 81); draw_num((1,4), 7); draw_num((2,4), 13); draw_num((3,4), 19); draw_num((4, 4), 25); draw_num((0, 4), 1); draw_num((1, 0), 33); draw_num((2, 0), 49); draw_num((3, 0), 65); void foo(int x, int y, string n) { label(n, (x+0.5,y+0.5), p = fontsize(19pt)); } foo(2, 4, " "); foo(3, 4, " "); foo(0, 3, " "); foo(2, 3, " "); foo(1, 2, " "); foo(3, 2, " "); foo(1, 1, " "); foo(2, 1, " "); foo(3, 1, " "); foo(4, 1, " "); foo(2, 0, " "); foo(3, 0, " "); foo(0, 1, " "); foo(0, 2, " "); foo(1, 0, " "); foo(1, 3, " "); foo(1, 4, " "); foo(3, 3, " "); foo(4, 2, " "); foo(4, 3, " "); [/asy]

We must find the third term of the arithmetic sequence with a first term of $13$ and a fifth term of $49$. The common difference of this sequence is $\frac{49-13}4=9$, so the third term is $13+2\cdot 9=\boxed{\textbf{(B) }31}$.

Solution 2

The middle term of the first row is $\frac{25+1}{2}=13$, since the middle number is just the average in an arithmetic sequence. Similarly, the middle of the bottom row is $\frac{17+81}{2}=49$. Applying this again for the middle column, the answer is $\frac{49+13}{2}=\boxed{\textbf{(B)}~31}$.

Solution 3

The value of $X$ is simply the average of the average values of both diagonals that contain $X$. This is $\frac{\frac{1+81}{2}+\frac{17+25}{2}}{2} =\frac{\frac{82}{2}+\frac{42}{2}}{2} = \frac{41+21}{2} = \boxed{\textbf{(B)}~31}$

Video Solution (HOW TO THINK CRITICALLY!!!)

https://youtu.be/1BUbtuK_MXI

~Education, the Study of Everything


Video Solution

https://youtu.be/tKsYSBdeVuw?t=1258

Video Solution

https://youtu.be/QFC4VmFa9Kc

~savannahsolver

See Also

2015 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 17 		Followed by
Problem 19
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_18&oldid=228035"