Difference between revisions of "2009 AIME I Problems/Problem 1"
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Consider the three-digit number <math>\overline{abc}</math>. If its digits form a geometric sequence, we must have that <math>{a \over b} = {b \over c}</math>, that is, <math>b^2 = ac</math>. | Consider the three-digit number <math>\overline{abc}</math>. If its digits form a geometric sequence, we must have that <math>{a \over b} = {b \over c}</math>, that is, <math>b^2 = ac</math>. | ||
| − | The minimum and maximum geometric numbers occur when <math>a</math> is | + | The minimum and maximum geometric numbers occur when <math>a</math> is minimized and maximized, respectively. The minimum occurs when <math>a = 1</math>; letting <math>b = 2</math> and <math>c = 4</math> achieves this, so the smallest possible geometric number is 124. |
For the maximum, we have that <math>b^2 = 9c</math>; <math>b</math> is maximized when <math>9c</math> is the greatest possible perfect square; this happens when <math>c = 4</math>, yielding <math>b = 6</math>. Thus, the largest possible geometric number is 964. | For the maximum, we have that <math>b^2 = 9c</math>; <math>b</math> is maximized when <math>9c</math> is the greatest possible perfect square; this happens when <math>c = 4</math>, yielding <math>b = 6</math>. Thus, the largest possible geometric number is 964. | ||
Revision as of 16:22, 4 May 2018
Problem
Call a 
-digit number geometric if it has 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 
for some integer 
, because a whole number should be attained for the 3rd term as well. When 
, the number is 
. When 
, the number is 
. When 
, we get 
, but the integers must be distinct. By the same logic, the smallest geometric number is 
. The largest geometric number is and the smallest is . Thus the difference is 
.
Solution 2
Consider the three-digit number 
. If its digits form a geometric sequence, we must have that 
, that is, 
.
The minimum and maximum geometric numbers occur when 
is minimized and maximized, respectively. The minimum occurs when 
; letting 
and 
achieves this, so the smallest possible geometric number is 124.
For the maximum, we have that 
; 
is maximized when 
is the greatest possible perfect square; this happens when , yielding 
. Thus, the largest possible geometric number is 964.
Our answer is thus .
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 
See also
| 2009 AIME I (Problems • Answer Key • Resources) | ||
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
