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

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== Problem ==
 
== Problem ==
  
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== Solution ==
 
== Solution ==
  
Assume that the largest geometric number starts with a nine. We know that the common ratio must be k/3, because a whole number should be attained for the 3rd term as well. When k = 1, the number is <math>931</math>. When k = 2, the number is 964. When k = 3, 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 1 ===
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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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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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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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== See also ==
 
== See also ==

Revision as of 04:53, 21 March 2009

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

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

Maybe an easier way how to see the solution: Consider a three-digit number $\overline{abc}$. If it is geometric, then we must have $\dfrac ab = \dfrac bc$, or equivalently $c = \dfrac{b^2}a$.

For $(a,b)=(9,8)$ we get $c=\dfrac{64}9$, which is not an integer. Similarly, for $(a,b)=(9,7)$ we will get a non-integer $c$. For $(a,b)=(9,6)$ we get $c=\dfrac{36}9 = 4$, hence $964$ is the largest three-digit geometric number. And as obviously the smallest possible pair $(a,b)=(1,2)$ provides the solution $124$, the answer is $964 - 124 = \boxed{840}$.


See also

2009 AIME I (ProblemsAnswer KeyResources)
Preceded by
First Question
Followed by
Problem 2
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All AIME Problems and Solutions
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