Difference between revisions of "2015 AMC 8 Problems/Problem 18"

Revision as of 15:43, 16 January 2021

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 $5\times5$ array is an arithmetic sequence with five terms. What is the value of $\text{X}$ ?

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

$[asy] size(3.85cm); label("$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 $\frac{\frac{1+81}{2}+\frac{17+25}{2}}{2} =\frac{\frac{82}{2}+\frac{42}{2}}{2} = \frac{41+21}{2} = \boxed{\textbf{(B)}~31}$

@@ Line 54: / Line 54: @@
 </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:
 <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>.
-==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>.
-==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>

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
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