2009 AIME I Problems/Problem 1

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

2009 AIME I (Problems • Answer Key • Resources)
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2009 AIME I Problems/Problem 1

Contents

Problem

Solution

Solution 1

Solution 2

See also