2015 AMC 8 Problems/Problem 18

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]

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 $\frac{\frac{1+81}{2}+\frac{17+25}{2}}{2} =\frac{\frac{82}{2}+\frac{42}{2}}{2} = \frac{41+21}{2} = \boxed{\textbf{(B)}~31}$

See Also

2015 AMC 8 (ProblemsAnswer KeyResources)
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

Invalid username
Login to AoPS