Difference between revisions of "2007 AIME II Problems/Problem 7"
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== Problem == | == Problem == | ||
− | Given a real number <math>x,</math> let <math>\lfloor x \rfloor</math> denote the greatest integer less than or equal to <math>x.</math> For a certain integer <math>k,</math> there are exactly <math>70</math> positive integers <math>n_{1}, n_{2}, \ldots, n_{70}</math> such that <math>k=\lfloor\sqrt[3]{n_{1}}\rfloor = \lfloor\sqrt[3]{n_{ | + | Given a [[real number]] <math>x,</math> let <math>\lfloor x \rfloor</math> denote the [[floor function|greatest integer]] less than or equal to <math>x.</math> For a certain [[integer]] <math>k,</math> there are exactly <math>70</math> positive integers <math>n_{1}, n_{2}, \ldots, n_{70}</math> such that <math>k=\lfloor\sqrt[3]{n_{1}}\rfloor = \lfloor\sqrt[3]{n_{2}}\rfloor = \cdots = \lfloor\sqrt[3]{n_{70}}\rfloor</math> and <math>k</math> divides <math>n_{i}</math> for all <math>i</math> such that <math>1 \leq i \leq 70.</math> |
Find the maximum value of <math>\frac{n_{i}}{k}</math> for <math>1\leq i \leq 70.</math> | Find the maximum value of <math>\frac{n_{i}}{k}</math> for <math>1\leq i \leq 70.</math> | ||
− | == Solution == | + | ==Solution== |
− | For <math>x = 1</math>, we see that <math>\sqrt[3]{1} \ldots \sqrt[3]{7}</math> all work, giving 7 integers. For <math>x=2</math>, we see that in <math>\sqrt[3]{8} \ldots \sqrt[3]{26}</math>, all of the even numbers work, giving 10 integers. For <math>x = 3</math>, we get 13, and so on. We can predict that at <math>x = 22</math> we get 70. | + | === Solution 1=== |
+ | For <math>x = 1</math>, we see that <math>\sqrt[3]{1} \ldots \sqrt[3]{7}</math> all work, giving 7 integers. For <math>x=2</math>, we see that in <math>\sqrt[3]{8} \ldots \sqrt[3]{26}</math>, all of the [[even]] numbers work, giving 10 integers. For <math>x = 3</math>, we get 13, and so on. We can predict that at <math>x = 22</math> we get 70. | ||
To prove this, note that all of the numbers from <math>\sqrt[3]{x^3} \ldots \sqrt[3]{(x+1)^3 - 1}</math> divisible by <math>x</math> work. Thus, <math>\frac{(x+1)^3 - 1 - x^3}{x} + 1 = \frac{3x^2 + 3x + 1 - 1}{x} + 1 = 3x + 4</math> (the one to be inclusive) integers will fit the conditions. <math>3k + 4 = 70 \Longrightarrow k = 22</math>. | To prove this, note that all of the numbers from <math>\sqrt[3]{x^3} \ldots \sqrt[3]{(x+1)^3 - 1}</math> divisible by <math>x</math> work. Thus, <math>\frac{(x+1)^3 - 1 - x^3}{x} + 1 = \frac{3x^2 + 3x + 1 - 1}{x} + 1 = 3x + 4</math> (the one to be inclusive) integers will fit the conditions. <math>3k + 4 = 70 \Longrightarrow k = 22</math>. | ||
− | The maximum value of <math> | + | The maximum value of <math>n_i = (x + 1)^3 - 1</math>. Therefore, the solution is <math>\frac{23^3 - 1}{22} = \frac{(23 - 1)(23^2 + 23 + 1)}{22} = 529 + 23 + 1 = 553</math>. |
+ | |||
+ | ===Solution 2=== | ||
+ | Obviously <math>k</math> is positive. Then, we can let <math>n_1</math> equal <math>k^3</math> and similarly let <math>n_i</math> equal <math>k^3 + (i - 1)k</math>. | ||
+ | |||
+ | The wording of this problem (which uses "exactly") tells us that <math>k^3+69k<(k+1)^3 = k^3 + 3k^2 + 3k+1 \leq k^3+70k</math>. Taking away <math>k^3</math> from our inequality results in <math>69k<3k^2+3k+1\leq 70k</math>. Since <math>69k</math>, <math>3k^2+3k+1</math>, and <math>70k</math> are all integers, this inequality is equivalent to <math>69k\leq 3k^2+3k<70k</math>. Since <math>k</math> is positive, we can divide the inequality by <math>k</math> to get <math>69 \leq 3k+3 < 70</math>. Clearly the only <math>k</math> that satisfies is <math>k=22</math>. | ||
+ | |||
+ | Then, <math>\frac{n_{70}}{k}=k^2+69=484+69=\boxed{553}</math> is the maximum value of <math>\frac{n_i}{k}</math>. (Remember we set <math>n_i</math> equal to <math>k^3 + (i - 1)k</math>!) | ||
+ | |||
+ | ==Video Solution by the Beauty of Math== | ||
+ | https://youtu.be/42kXJgD_b-A | ||
== See also == | == See also == | ||
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[[Category:Intermediate Algebra Problems]] | [[Category:Intermediate Algebra Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 18:04, 4 December 2020
Contents
Problem
Given a real number let
denote the greatest integer less than or equal to
For a certain integer
there are exactly
positive integers
such that
and
divides
for all
such that
Find the maximum value of for
Solution
Solution 1
For , we see that
all work, giving 7 integers. For
, we see that in
, all of the even numbers work, giving 10 integers. For
, we get 13, and so on. We can predict that at
we get 70.
To prove this, note that all of the numbers from divisible by
work. Thus,
(the one to be inclusive) integers will fit the conditions.
.
The maximum value of . Therefore, the solution is
.
Solution 2
Obviously is positive. Then, we can let
equal
and similarly let
equal
.
The wording of this problem (which uses "exactly") tells us that . Taking away
from our inequality results in
. Since
,
, and
are all integers, this inequality is equivalent to
. Since
is positive, we can divide the inequality by
to get
. Clearly the only
that satisfies is
.
Then, is the maximum value of
. (Remember we set
equal to
!)
Video Solution by the Beauty of Math
See also
2007 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 6 |
Followed by Problem 8 | |
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.