2009 AIME I Problems/Problem 1

Revision as of 16:00, 21 March 2009 by Zhero (talk | contribs) (Solution 2)

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

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 minized 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 $964 - 124 = \boxed{840}$.

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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