Difference between revisions of "1989 AIME Problems/Problem 4"
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== Solution == | == Solution == | ||
− | Since the middle term of an [[arithmetic progression]] with an odd number of terms is the average of the series, we know <math>b + c + d = 3c</math> and <math>a + b + c + d + e = 5c</math>. Thus, <math>c</math> must be in the form of <math>3 \cdot x^2</math> based upon the first part and in the form of <math>5^2 \cdot y^3</math> based upon the second part, with <math>x</math> and <math>y</math> denoting an [[integer]]s. <math>c</math> is minimized if it’s [[prime factorization]] contains only <math>3,5</math>, and since there is a cubed term in <math>5^2 \cdot y^3</math>, <math>3^3</math> must be a factor of <math>c</math>. <math>3^35^2 = 675</math>, which works as the solution. | + | Since the middle term of an [[arithmetic progression]] with an odd number of terms is the average of the series, we know <math>b + c + d = 3c</math> and <math>a + b + c + d + e = 5c</math>. Thus, <math>c</math> must be in the form of <math>3 \cdot x^2</math> based upon the first part and in the form of <math>5^2 \cdot y^3</math> based upon the second part, with <math>x</math> and <math>y</math> denoting an [[integer]]s. <math>c</math> is minimized if it’s [[prime factorization]] contains only <math>3,5</math>, and since there is a cubed term in <math>5^2 \cdot y^3</math>, <math>3^3</math> must be a factor of <math>c</math>. <math>3^35^2 = \boxed{675}</math>, which works as the solution. |
== See also == | == See also == | ||
{{AIME box|year=1989|num-b=3|num-a=5}} | {{AIME box|year=1989|num-b=3|num-a=5}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 23:40, 21 October 2018
Problem
If are consecutive positive integers such that is a perfect square and is a perfect cube, what is the smallest possible value of ?
Solution
Since the middle term of an arithmetic progression with an odd number of terms is the average of the series, we know and . Thus, must be in the form of based upon the first part and in the form of based upon the second part, with and denoting an integers. is minimized if it’s prime factorization contains only , and since there is a cubed term in , must be a factor of . , which works as the solution.
See also
1989 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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