2009 AIME I Problems/Problem 1

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

2009 AIME I (Problems • Resources)
Preceded by First Question	Followed by Problem 2
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

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