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. |
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+ | |||
+ | ==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 | ||
+ | |||
+ | |||
+ | == Solution 3 == | ||
+ | |||
+ | |||
+ | Let the numbers be <math>a,a+1,a+2,a+3,a+4.</math> When then know <math>3a+6</math> is a perfect cube and <math>5a+10</math> is perfect cube. Since <math>5a+10</math> is divisible by <math>5</math> we know that <math>5a+10 = (5k)^3</math> since otherwise we get a contradiction. This means <math>a = 25k^3 - 2</math> in which plugging into the other expression we know <math>3(25k^3 - 2) + 6 = 75k^3</math> is a perfect square. We know <math>75 = 5^2 \cdot 3</math> so we let <math>k = 3</math> to obtain the perfect square. This means that <math>c = a+2 = (25 \cdot 27 - 2)+2 = 25 \cdot 27 = \boxed{675}.</math> | ||
== 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}} |
Latest revision as of 16:13, 14 September 2020
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
Solution 3
Let the numbers be When then know is a perfect cube and is perfect cube. Since is divisible by we know that since otherwise we get a contradiction. This means in which plugging into the other expression we know is a perfect square. We know so we let to obtain the perfect square. This means that
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.