Difference between revisions of "2009 AIME I Problems/Problem 1"
Pi is 3.14 (talk | contribs) (→Video Solution) |
|||
| (2 intermediate revisions by one other user not shown) | |||
| Line 5: | Line 5: | ||
== Solution 1 == | == 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 <math>k/3</math> for some integer <math>k</math>, because a whole number should be attained for the 3rd term as well. When <math>k = 1</math>, the number is <math>931</math>. When <math>k = 2</math>, the number is <math>964</math>. When <math>k = 3</math>, we get <math>999</math>, but the integers must be distinct. By the same logic, the smallest geometric number is <math>124</math>. The largest geometric number is <math>964</math> and the smallest is <math>124</math>. Thus the difference is <math>964 - 124 = \boxed{840}</math>. | + | Assume that the largest geometric number starts with a <math>9</math>. We know that the common ratio must be a rational of the form <math>k/3</math> for some integer <math>k</math>, because a whole number should be attained for the 3rd term as well. When <math>k = 1</math>, the number is <math>931</math>. When <math>k = 2</math>, the number is <math>964</math>. When <math>k = 3</math>, we get <math>999</math>, but the integers must be distinct. By the same logic, the smallest geometric number is <math>124</math>. The largest geometric number is <math>964</math> and the smallest is <math>124</math>. Thus the difference is <math>964 - 124 = \boxed{840}</math>. |
== Solution 2 == | == Solution 2 == | ||
| − | Consider the three-digit number abc. If its digits form a geometric progression, 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>abc</math>. If its digits form a geometric progression, 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 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. | 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. | ||
| Line 18: | Line 18: | ||
==Solution 3== | ==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 <math>\boxed{840}.</math> | + | The smallest geometric number is <math>124</math> because <math>123</math> and any number containing a zero does not work. <math>964</math> is the largest geometric number because the middle digit cannot be 8 or 7. Subtracting the numbers gives <math>\boxed{840}.</math> |
| + | |||
| + | == Video Solution by OmegaLearn == | ||
| + | https://youtu.be/1-iWPCWPsLw?t=195 | ||
| + | |||
| + | ~ pi_is_3.14 | ||
==Video Solution== | ==Video Solution== | ||
| Line 26: | Line 31: | ||
==Video Solution 2== | ==Video Solution 2== | ||
| − | https:// | + | https://www.youtube.com/watch?v=P00iOJdQiL4 |
~Shreyas S | ~Shreyas S | ||
Latest revision as of 02:32, 16 January 2023
Contents
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 1
Assume that the largest geometric number starts with a 
. 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 progression, 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 because 
and any number containing a zero does not work. is the largest geometric number because the middle digit cannot be 8 or 7. Subtracting the numbers gives 
Video Solution by OmegaLearn
https://youtu.be/1-iWPCWPsLw?t=195
~ pi_is_3.14
Video Solution
~IceMatrix
Video Solution 2
https://www.youtube.com/watch?v=P00iOJdQiL4
~Shreyas S
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. 
