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.
  
== Solution ==
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== Solution 1 ==
  
=== Solution 1 ===
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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>.
  
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 ==
  
=== Solution 2 ===
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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>.
  
Maybe an easier way how to see the solution: Consider a three-digit number <math>\overline{abc}</math>. If it is geometric, then we must have <math>\dfrac ab = \dfrac bc</math>, or equivalently <math>c = \dfrac{b^2}a</math>.
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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.  
  
For <math>(a,b)=(9,8)</math> we get <math>c=\dfrac{64}9</math>, which is not an integer. Similarly, for <math>(a,b)=(9,7)</math> we will get a non-integer <math>c</math>. For <math>(a,b)=(9,6)</math> we get <math>c=\dfrac{36}9 = 4</math>, hence <math>964</math> is the largest three-digit geometric number. And as obviously the smallest possible pair <math>(a,b)=(1,2)</math> provides the solution <math>124</math>, the answer is <math>964 - 124 = \boxed{840}</math>.
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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.  
  
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Our answer is thus <math>964 - 124 = \boxed{840}</math>.
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==Solution 3==
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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>
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==Video Solution==
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https://youtu.be/NL79UexadzE
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~IceMatrix
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==Video Solution 2==
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https://www.youtube.com/watch?v=P00iOJdQiL4
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~Shreyas S
  
 
== See also ==
 
== See also ==
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[[Category:Intermediate Number Theory Problems]]
 
[[Category:Intermediate Number Theory Problems]]
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{{MAA Notice}}

Latest revision as of 15:47, 22 August 2020

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

https://youtu.be/NL79UexadzE

~IceMatrix

Video Solution 2

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

~Shreyas S

See also

2009 AIME I (ProblemsAnswer KeyResources)
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