Difference between revisions of "1989 AIME Problems/Problem 4"
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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> | 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> | ||
+ | |||
+ | == Solution 4 == | ||
+ | |||
+ | |||
+ | (This is literally a combination of 1 and 3) | ||
+ | |||
+ | Since <math>a</math>, <math>b</math>, <math>c</math>, <math>d</math>, and <math>e</math> are consecutive, <math>a = c-2</math>, <math>b = c-1</math>, <math>c=c</math>, <math>d = c+1</math>, and <math>e = c+2</math>. | ||
+ | |||
+ | Because <math>b+c+d = 3c</math> is a perfect square, and <math>a+b+c+d+e = 5c</math> is a perfect cube, we can express <math>c</math> as <math>c = 3^{n} \cdot 5^{k}</math>. | ||
+ | |||
+ | Now, by the problem's given information, | ||
+ | |||
+ | <math>k \equiv 0 \text{(mod 2)}</math> | ||
+ | |||
+ | <math>n \equiv 0 \text{(mod 3)}</math> | ||
+ | |||
+ | and because ALL exponents have to be cubes/squares, | ||
+ | |||
+ | <math>k \equiv 2 \text{(mod 3)}</math> | ||
+ | |||
+ | <math>n \equiv 1 \text{(mod 2)}</math> | ||
+ | |||
+ | Therefore, | ||
+ | |||
+ | <math>k = 2</math> | ||
+ | |||
+ | <math>n = 3</math> | ||
+ | |||
+ | <math>c = 3^3 \cdot 5^2 = \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}} |
Revision as of 12:40, 3 March 2024
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
Solution 4
(This is literally a combination of 1 and 3)
Since ,
,
,
, and
are consecutive,
,
,
,
, and
.
Because is a perfect square, and
is a perfect cube, we can express
as
.
Now, by the problem's given information,
and because ALL exponents have to be cubes/squares,
Therefore,
.
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.