Difference between revisions of "2009 AIME I Problems/Problem 1"
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== Problem == | == Problem == | ||
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== Solution == | == Solution == | ||
− | Assume that the largest geometric number starts with a nine. We know that the common ratio must be k/3, because a whole number should be attained for the 3rd term as well. When k = 1, the number is <math>931</math>. When k = 2, the number is 964. When k = 3, 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 1 === |
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+ | Assume that the largest geometric number starts with a nine. 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>. | ||
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+ | === Solution 2 === | ||
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+ | Maybe an easier way how to see the solution: Consider a three-digit number <math>\overline{abc}</math>. If it is geometric, then we must have <math>\dfrac ab = \dfrac bc</math>, or equivalently <math>c = \dfrac{b^2}a</math>. | ||
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+ | For <math>(a,b)=(9,8)</math> we get <math>c=\dfrac{64}9</math>, which is not an integer. Similarly, for <math>(a,b)=(9,7)</math> we will get a non-integer <math>c</math>. For <math>(a,b)=(9,6)</math> we get <math>c=\dfrac{36}9 = 4</math>, hence <math>964</math> is the largest three-digit geometric number. And as obviously the smallest possible pair <math>(a,b)=(1,2)</math> provides the solution <math>124</math>, the answer is <math>964 - 124 = \boxed{840}</math>. | ||
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== See also == | == See also == |
Revision as of 04:53, 21 March 2009
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
Maybe an easier way how to see the solution: Consider a three-digit number . If it is geometric, then we must have , or equivalently .
For we get , which is not an integer. Similarly, for we will get a non-integer . For we get , hence is the largest three-digit geometric number. And as obviously the smallest possible pair provides the solution , the answer is .
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 |