AIME 2009 I · 第 1 题

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

Video Solution 3

https://www.youtube.com/watch?v=wT8PRBqD-Ds