Difference between revisions of "1985 AIME Problems/Problem 10"
m |
(solution added) |
||
| Line 6: | Line 6: | ||
where <math>x</math> is a [[real number]], and <math>\lfloor z \rfloor</math> denotes the greatest [[integer]] less than or equal to <math>z</math>? | where <math>x</math> is a [[real number]], and <math>\lfloor z \rfloor</math> denotes the greatest [[integer]] less than or equal to <math>z</math>? | ||
== Solution == | == Solution == | ||
| − | {{ | + | We will be able to reach the same number of integers while <math>x</math> ranges from 0 to 1 as we will when <math>x</math> ranges from <math>n</math> to <math>n + 1</math> for any integer <math>n</math>. Since <math>\lfloor 2\cdot50 \rfloor + \lfloor 4\cdot50 \rfloor + \lfloor 6\cdot50 \rfloor + \lfloor 8\cdot50 \rfloor = 100 + 200 + 300 + 400</math>, the answer must be exactly 50 times the number of integers we will be able to reach as <math>x</math> ranges from 0 to 1, including 1 but excluding 0. |
| + | |||
| + | As we change the value of <math>x</math>, the value of our [[expression]] changes only when <math>x</math> crosses [[rational number]] of the form <math>\frac{m}{n}</math>, where <math>n</math> is divisible by 2, 4, 6 or 8. Thus, we need only see what happens at the numbers of the form <math>\frac{m}{\textrm{gcd}(2, 4, 6, 8)} = \frac{m}{24}</math>. This gives us 24 calculations to make; we summarize the results here: | ||
| + | |||
| + | <math>\frac{1}{24}, \frac{2}{24} \to 0</math> | ||
| + | |||
| + | <math>\frac{3}{24} \to 1</math> | ||
| + | |||
| + | <math>\frac{4}{24}, \frac{5}{24} \to 2</math> | ||
| + | |||
| + | <math>\frac{6}{24}, \frac{7}{24} \to 4</math> | ||
| + | |||
| + | <math>\frac{8}{24} \to 5</math> | ||
| + | |||
| + | <math>\frac{9}{24}, \frac{10}{24}, \frac{11}{24} \to 6</math> | ||
| + | |||
| + | <math>\frac{12}{24}, \frac{13}{24}, \frac{14}{24} \to 10</math> | ||
| + | |||
| + | <math>\frac{15}{24} \to 11</math> | ||
| + | |||
| + | <math>\frac{16}{24},\frac{17}{24} \to 12</math> | ||
| + | |||
| + | <math>\frac{18}{24}, \frac{19}{24} \to 14</math> | ||
| + | |||
| + | <math>\frac{20}{24}\to 15</math> | ||
| + | |||
| + | <math>\frac{21}{24}, \frac{22}{24}, \frac{23}{24} \to16</math> | ||
| + | |||
| + | <math>\frac{24}{24} \to 20</math> | ||
| + | |||
| + | Thus, we hit 12 of the first 20 integers and so we hit <math>50 \cdot 12 = 600</math> of the first 100. | ||
| + | |||
== See also == | == See also == | ||
* [[1985 AIME Problems/Problem 9 | Previous problem]] | * [[1985 AIME Problems/Problem 9 | Previous problem]] | ||
| Line 12: | Line 43: | ||
* [[1985 AIME Problems]] | * [[1985 AIME Problems]] | ||
* [[Floor function]] | * [[Floor function]] | ||
| + | |||
| + | [[Category:Intermediate Algebra Problems]] | ||
Revision as of 17:48, 22 January 2007
Problem
How many of the first 1000 positive integers can be expressed in the form
,
where 
is a real number, and 
denotes the greatest integer less than or equal to 
?
Solution
We will be able to reach the same number of integers while ranges from 0 to 1 as we will when ranges from 
to 
for any integer . Since 
, the answer must be exactly 50 times the number of integers we will be able to reach as ranges from 0 to 1, including 1 but excluding 0.
As we change the value of , the value of our expression changes only when crosses rational number of the form 
, where is divisible by 2, 4, 6 or 8. Thus, we need only see what happens at the numbers of the form 
. This gives us 24 calculations to make; we summarize the results here:
Thus, we hit 12 of the first 20 integers and so we hit 
of the first 100.
