Difference between revisions of "2007 AIME I Problems/Problem 1"
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The [[prime factorization]] of <math>24</math> is <math>2^3\cdot3</math>. Thus, each square must have at least <math>3</math> factors of <math>2</math> and <math>1</math> factor of <math>3</math> and its square root must have <math>2</math> factors of <math>2</math> and <math>1</math> factor of <math>3</math>. | The [[prime factorization]] of <math>24</math> is <math>2^3\cdot3</math>. Thus, each square must have at least <math>3</math> factors of <math>2</math> and <math>1</math> factor of <math>3</math> and its square root must have <math>2</math> factors of <math>2</math> and <math>1</math> factor of <math>3</math>. | ||
This means that each square is in the form <math>(12c)^2</math>, where <math>12 c</math> is a positive integer less than <math>\sqrt{10^6}</math>. There are <math>\left\lfloor \frac{1000}{12}\right\rfloor = \boxed{083}</math> solutions. | This means that each square is in the form <math>(12c)^2</math>, where <math>12 c</math> is a positive integer less than <math>\sqrt{10^6}</math>. There are <math>\left\lfloor \frac{1000}{12}\right\rfloor = \boxed{083}</math> solutions. | ||
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+ | == Solution 2 == | ||
+ | The perfect squares divisible by <math>24</math> are all multiples of <math>12</math>: <math>12^2</math>, <math>24^2</math>, <math>36^2</math>, <math>48^2</math>... Since they all have to be less than <math>10^6</math>, or <math>1000^2</math>, the closest multiple of <math>12</math> to <math>1000</math> is <math>996</math> (<math>12*83</math>), so we know that this is the last term in the sequence. Therefore, we know that there are <math>\boxed{083}</math> perfect squares divisible by <math>24</math> that are less than <math>10^6</math>. | ||
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== See also == | == See also == | ||
{{AIME box|year=2007|n=I|before=First Question|num-a=2}} | {{AIME box|year=2007|n=I|before=First Question|num-a=2}} |
Revision as of 13:54, 7 February 2023
Contents
Problem
How many positive perfect squares less than are multiples of
?
Solution
The prime factorization of is
. Thus, each square must have at least
factors of
and
factor of
and its square root must have
factors of
and
factor of
.
This means that each square is in the form
, where
is a positive integer less than
. There are
solutions.
Solution 2
The perfect squares divisible by are all multiples of
:
,
,
,
... Since they all have to be less than
, or
, the closest multiple of
to
is
(
), so we know that this is the last term in the sequence. Therefore, we know that there are
perfect squares divisible by
that are less than
.
See also
2007 AIME I (Problems • Answer Key • Resources) | ||
Preceded by First Question |
Followed by Problem 2 | |
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.