Difference between revisions of "1980 AHSME Problems/Problem 25"
(Created page with "== Problem == In the non-decreasing sequence of odd integers <math>\{a_1,a_2,a_3,\ldots \}=\{1,3,3,3,5,5,5,5,5,\ldots \}</math> each odd positive integer <math>k</math> appears ...") |
(→Solution) |
||
| Line 13: | Line 13: | ||
== Solution == | == Solution == | ||
| − | <math>\ | + | Solution by e_power_pi_times_i |
| + | |||
| + | Because the set consists of odd numbers, and since <math>\lfloor{}\sqrt{n+c}\rfloor{}</math> is an integer and can be odd or even, <math>b = 2</math> and <math>|a| = 1</math>. However, given that <math>\lfloor{}\sqrt{n+c}\rfloor{}</math> can be <math>0</math>, <math>a = 1</math>. Then, <math>a_1 = 1 = 2\lfloor{}\sqrt{1+c}\rfloor{}+1</math>, and <math>\lfloor{}\sqrt{1+c}\rfloor{}</math> = 0, and <math>c = -1</math> because <math>c</math> is an integer. <math>b+c+d = 2+(-1)+1 = \boxed{\text{(C)}\ 2}</math> | ||
== See also == | == See also == | ||
Latest revision as of 12:00, 17 November 2016
Problem
In the non-decreasing sequence of odd integers 
each odd positive integer 
appears times. It is a fact that there are integers 
, and 
such that for all positive integers 
,
,
where 
denotes the largest integer not exceeding 
. The sum 
equals
Solution
Solution by e_power_pi_times_i
Because the set consists of odd numbers, and since 
is an integer and can be odd or even, 
and 
. However, given that can be 
, 
. Then, 
, and 
= 0, and 
because 
is an integer. 
See also
| 1980 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 24 |
Followed by Problem 26 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
