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 = \boxed{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. | ||
+ | |||
+ | |||
+ | ==Solution 2== | ||
+ | Let <math>b</math>, <math>c</math>, <math>d</math>, and <math>e</math> equal <math>a+1</math>, <math>a+2</math>, <math>a+3</math>, and <math>a+4</math>, respectively. Call the square and cube <math>k^2</math> and <math>m^3</math>, where both k and m are integers. Then: | ||
+ | |||
+ | <math>5a + 10 = m^3</math> | ||
+ | |||
+ | Now we know <math>m^3</math> is a multiple of 125 and <math>m</math> is a multiple of 5. The lower <math>m</math> is, the lower the value of <math>c</math> will be. Start from 5 and add 5 each time. | ||
+ | |||
+ | <math>m = 5</math> gives no solution for k | ||
+ | |||
+ | <math>m = 10</math> gives no solution for k | ||
+ | |||
+ | <math>m = 15</math> gives a solution for k. | ||
+ | |||
+ | |||
+ | <math>10 + 5a = 15^3</math> | ||
+ | |||
+ | |||
+ | <math>2 + a = 675</math> | ||
+ | |||
+ | |||
+ | <math>c = \boxed{675}</math> | ||
+ | |||
+ | |||
+ | -jackshi2006 | ||
== 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 12:59, 30 August 2020
Contents
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.
Solution 2
Let , , , and equal , , , and , respectively. Call the square and cube and , where both k and m are integers. Then:
Now we know is a multiple of 125 and is a multiple of 5. The lower is, the lower the value of will be. Start from 5 and add 5 each time.
gives no solution for k
gives no solution for k
gives a solution for k.
-jackshi2006
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.