2009 AIME I Problems/Problem 1

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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 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
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions