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 "2009 AIME I Problems/Problem 1"
(→‎Video Solution)
 
(7 intermediate revisions by 2 users not shown)
Line 5: Line 5:
 
== Solution 1 ==
 
== Solution 1 ==
  
−
Assume that the largest geometric number starts with a nine. We know that the common ratio must be a rational of the form <math>k/3</math> for some integer <math>k</math>, because a whole number should be attained for the 3rd term as well. When <math>k = 1</math>, the number is <math>931</math>. When <math>k = 2</math>, the number is <math>964</math>. When <math>k = 3</math>, we get <math>999</math>, but the integers must be distinct. By the same logic, the smallest geometric number is <math>124</math>. The largest geometric number is <math>964</math> and the smallest is <math>124</math>. Thus the difference is <math>964 - 124 = \boxed{840}</math>.
+
Assume that the largest geometric number starts with a <math>9</math>. We know that the common ratio must be a rational of the form <math>k/3</math> for some integer <math>k</math>, because a whole number should be attained for the 3rd term as well. When <math>k = 1</math>, the number is <math>931</math>. When <math>k = 2</math>, the number is <math>964</math>. When <math>k = 3</math>, we get <math>999</math>, but the integers must be distinct. By the same logic, the smallest geometric number is <math>124</math>. The largest geometric number is <math>964</math> and the smallest is <math>124</math>. Thus the difference is <math>964 - 124 = \boxed{840}</math>.
  
 
== Solution 2 ==
 
== Solution 2 ==
  
−
Consider the three-digit number <math>\overline{abc}</math>. If its digits form a geometric sequence, we must have that <math>{a \over b} = {b \over c}</math>, that is, <math>b^2 = ac</math>.  
+
Consider the three-digit number <math>abc</math>. If its digits form a geometric progression, we must have that <math>{a \over b} = {b \over c}</math>, that is, <math>b^2 = ac</math>.  
  
 
The minimum and maximum geometric numbers occur when <math>a</math> is minimized and maximized, respectively. The minimum occurs when <math>a = 1</math>; letting <math>b = 2</math> and <math>c = 4</math> achieves this, so the smallest possible geometric number is 124.  
 
The minimum and maximum geometric numbers occur when <math>a</math> is minimized and maximized, respectively. The minimum occurs when <math>a = 1</math>; letting <math>b = 2</math> and <math>c = 4</math> achieves this, so the smallest possible geometric number is 124.  
Line 15: Line 15:
 
For the maximum, we have that <math>b^2 = 9c</math>; <math>b</math> is maximized when <math>9c</math> is the greatest possible perfect square; this happens when <math>c = 4</math>, yielding <math>b = 6</math>. Thus, the largest possible geometric number is 964.  
 
For the maximum, we have that <math>b^2 = 9c</math>; <math>b</math> is maximized when <math>9c</math> is the greatest possible perfect square; this happens when <math>c = 4</math>, yielding <math>b = 6</math>. Thus, the largest possible geometric number is 964.  
  
−
Our answer is thus <math>964 - 124 = 840.
+
Our answer is thus <math>964 - 124 = \boxed{840}</math>.
  
 
==Solution 3==
 
==Solution 3==
−
The smallest geometric number is 124 because 123 and any number containing a zero does not work. 964 is the largest geometric number because the middle digit cannot be 8 or 7. Subtracting the numbers gives </math>\boxed{840}.$
+
The smallest geometric number is <math>124</math> because <math>123</math> and any number containing a zero does not work. <math>964</math> is the largest geometric number because the middle digit cannot be 8 or 7. Subtracting the numbers gives <math>\boxed{840}.</math>
 +
 
 +
== Video Solution by OmegaLearn ==
 +
https://youtu.be/1-iWPCWPsLw?t=195
 +
 
 +
~ pi_is_3.14
  
 
==Video Solution==
 
==Video Solution==
Line 24: Line 29:
  
 
~IceMatrix
 
~IceMatrix
 +
 +
==Video Solution 2==
 +
https://www.youtube.com/watch?v=P00iOJdQiL4
 +
 +
~Shreyas S
  
 
== See also ==
 
== See also ==

Latest revision as of 02:32, 16 January 2023

Problem

Call a $3$-digit number geometric if it has $3$ distinct digits which, when read from left to right, form a geometric sequence. Find the difference between the largest and smallest geometric numbers.

Solution 1

Assume that the largest geometric number starts with a $9$. We know that the common ratio must be a rational of the form $k/3$ for some integer $k$, because a whole number should be attained for the 3rd term as well. When $k = 1$, the number is $931$. When $k = 2$, the number is $964$. When $k = 3$, we get $999$, but the integers must be distinct. By the same logic, the smallest geometric number is $124$. The largest geometric number is $964$ and the smallest is $124$. Thus the difference is $964 - 124 = \boxed{840}$.

Solution 2

Consider the three-digit number $abc$. If its digits form a geometric progression, we must have that ${a \over b} = {b \over c}$, that is, $b^2 = ac$.

The minimum and maximum geometric numbers occur when $a$ is minimized and maximized, respectively. The minimum occurs when $a = 1$; letting $b = 2$ and $c = 4$ achieves this, so the smallest possible geometric number is 124.

For the maximum, we have that $b^2 = 9c$; $b$ is maximized when $9c$ is the greatest possible perfect square; this happens when $c = 4$, yielding $b = 6$. Thus, the largest possible geometric number is 964.

Our answer is thus $964 - 124 = \boxed{840}$.

Solution 3

The smallest geometric number is $124$ because $123$ and any number containing a zero does not work. $964$ is the largest geometric number because the middle digit cannot be 8 or 7. Subtracting the numbers gives $\boxed{840}.$

Video Solution by OmegaLearn

https://youtu.be/1-iWPCWPsLw?t=195

~ pi_is_3.14

Video Solution

https://youtu.be/NL79UexadzE

~IceMatrix

Video Solution 2

https://www.youtube.com/watch?v=P00iOJdQiL4

~Shreyas S

See also

2009 AIME I (Problems • Answer Key • Resources)
Preceded by
First Question 		Followed by
Problem 2
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME 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=2009_AIME_I_Problems/Problem_1&oldid=186741"