Difference between revisions of "2009 AIME I Problems/Problem 1"
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Call a <math>3</math>-digit number ''geometric'' if it has <math>3</math> distinct digits which, when read from left to right, form a geometric sequence. Find the difference between the largest and smallest geometric numbers. | Call a <math>3</math>-digit number ''geometric'' if it has <math>3</math> 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 <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 == | |
− | = | + | 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. | |
− | For <math> | + | 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. |
+ | Our answer is thus <math>964 - 124 = \boxed{840}</math>. | ||
+ | |||
+ | ==Solution 3== | ||
+ | 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== | ||
+ | https://youtu.be/NL79UexadzE | ||
+ | |||
+ | ~IceMatrix | ||
+ | |||
+ | ==Video Solution 2== | ||
+ | https://www.youtube.com/watch?v=P00iOJdQiL4 | ||
+ | |||
+ | ~Shreyas S | ||
== See also == | == See also == | ||
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[[Category:Intermediate Number Theory Problems]] | [[Category:Intermediate Number Theory Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 15:47, 22 August 2020
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
~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 |
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