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Difference between revisions of "2009 AIME I Problems/Problem 1"

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Call a <math>3</math>-digit number ''geometric'' if it has <math>3</math> distinct digits which, when read from left to right, form a geometric sequence.  Find the difference between the largest and smallest geometric numbers.
 
Call a <math>3</math>-digit number ''geometric'' if it has <math>3</math> distinct digits which, when read from left to right, form a geometric sequence.  Find the difference between the largest and smallest geometric numbers.
  
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=== Solution 1 ===
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== 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 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>.
  
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=== Solution 2 ===
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== 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>\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>.  
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Our answer is thus <math>964 - 124 = \boxed{840}</math>.
 
Our answer is thus <math>964 - 124 = \boxed{840}</math>.
  
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===Solution 3===
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==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}.</math>
 
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}.</math>
  

Revision as of 12:56, 6 June 2019

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 nine. 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 $\overline{abc}$. If its digits form a geometric sequence, 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 $\boxed{840}.$

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

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