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 "2006 AMC 8 Problems/Problem 22"

(→‎Solution)
(→‎Video Solution)
 
(11 intermediate revisions by 7 users not shown)
Line 1: Line 1:
 
==Problem==
 
==Problem==
  
−
Three different one-digit positive integers are placed in the bottom row of cells. Numbers in adjacent cells are added and the sum is placed in the cell above them. In the second row, continue the same process to obtain a number in the top cell what is the difference between the largest and the smallest number possible in the top cell?
+
Three different one-digit positive integers are placed in the bottom row of cells. Numbers in adjacent cells are added and the sum is placed in the cell above them. In the second row, continue the same process to obtain a number in the top cell. What is the difference between the largest and smallest numbers possible in the top cell?
  
 +
<asy>
 +
path cell=((0,0)--(1,0)--(1,1)--(0,1)--cycle);
 +
path sw=((0,0)--(1,sqrt(3)));
 +
path se=((5,0)--(4,sqrt(3)));
 +
draw(cell, linewidth(1));
 +
draw(shift(2,0)*cell, linewidth(1));
 +
draw(shift(4,0)*cell, linewidth(1));
 +
draw(shift(1,3)*cell, linewidth(1));
 +
draw(shift(3,3)*cell, linewidth(1));
 +
draw(shift(2,6)*cell, linewidth(1));
 +
draw(shift(0.45,1.125)*sw, EndArrow);
 +
draw(shift(2.45,1.125)*sw, EndArrow);
 +
draw(shift(1.45,4.125)*sw, EndArrow);
 +
draw(shift(-0.45,1.125)*se, EndArrow);
 +
draw(shift(-2.45,1.125)*se, EndArrow);
 +
draw(shift(-1.45,4.125)*se, EndArrow);
 +
label("$+$", (1.5,1.5));
 +
label("$+$", (3.5,1.5));
 +
label("$+$", (2.5,4.5));</asy>
 +
 +
<math> \textbf{(A)}\ 16\qquad\textbf{(B)}\ 24\qquad\textbf{(C)}\ 25\qquad\textbf{(D)}\ 26\qquad\textbf{(E)}\ 35 </math>
  
 
==Solution==
 
==Solution==
  
−
('''D''') If the lower cells contain A, B and C, then the second row will contain
+
If the lower cells contain <math>A, B</math> and <math>C</math>, then the second row will contain
−
A + B and B + C, and the top cell will contain A + 2B + C. To obtain the
+
<math>A + B</math> and <math>B + C</math>, and the top cell will contain <math>A + 2B + C</math>. To obtain the
−
smallest sum, place 1 in the center cell and 2 and 3 in the outer ones. The top
+
smallest sum, place <math>1</math> in the center cell and <math>2</math> and <math>3</math> in the outer ones. The top
−
number will be 7. For the largest sum, place 9 in the center cell and 7 and 8 in
+
number will be <math>7</math>. For the largest sum, place <math>9</math> in the center cell and <math>7</math> and <math>8</math> in
−
the outer ones. This top number will be 33. The difference is 33 - 7 = 26.
+
the outer ones. This top number will be <math>33</math>. The difference is <math>33 - 7 = \boxed{\textbf{(D)}\ 26
 +
}</math>.
 +
 
 +
 
 +
==Video Solution==
 +
 
 +
https://www.youtube.com/watch?v=_FAL-t93VIk  ~David
 +
 
 +
==See Also==
 +
{{AMC8 box|year=2006|n=II|num-b=21|num-a=23}}
 +
{{MAA Notice}}

Latest revision as of 08:19, 18 June 2024

Problem

Three different one-digit positive integers are placed in the bottom row of cells. Numbers in adjacent cells are added and the sum is placed in the cell above them. In the second row, continue the same process to obtain a number in the top cell. What is the difference between the largest and smallest numbers possible in the top cell?

[asy] path cell=((0,0)--(1,0)--(1,1)--(0,1)--cycle); path sw=((0,0)--(1,sqrt(3))); path se=((5,0)--(4,sqrt(3))); draw(cell, linewidth(1)); draw(shift(2,0)*cell, linewidth(1)); draw(shift(4,0)*cell, linewidth(1)); draw(shift(1,3)*cell, linewidth(1)); draw(shift(3,3)*cell, linewidth(1)); draw(shift(2,6)*cell, linewidth(1)); draw(shift(0.45,1.125)*sw, EndArrow); draw(shift(2.45,1.125)*sw, EndArrow); draw(shift(1.45,4.125)*sw, EndArrow); draw(shift(-0.45,1.125)*se, EndArrow); draw(shift(-2.45,1.125)*se, EndArrow); draw(shift(-1.45,4.125)*se, EndArrow); label("$+$", (1.5,1.5)); label("$+$", (3.5,1.5)); label("$+$", (2.5,4.5));[/asy]

$\textbf{(A)}\ 16\qquad\textbf{(B)}\ 24\qquad\textbf{(C)}\ 25\qquad\textbf{(D)}\ 26\qquad\textbf{(E)}\ 35$

Solution

If the lower cells contain $A, B$ and $C$, then the second row will contain $A + B$ and $B + C$, and the top cell will contain $A + 2B + C$. To obtain the smallest sum, place $1$ in the center cell and $2$ and $3$ in the outer ones. The top number will be $7$. For the largest sum, place $9$ in the center cell and $7$ and $8$ in the outer ones. This top number will be $33$. The difference is $33 - 7 = \boxed{\textbf{(D)}\ 26 }$.


Video Solution

https://www.youtube.com/watch?v=_FAL-t93VIk ~David

See Also

2006 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 21 		Followed by
Problem 23
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=2006_AMC_8_Problems/Problem_22&oldid=220705"