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 ...")
 
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== Solution ==
 
== Solution ==
<math>\fbox{}</math>
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Solution by e_power_pi_times_i
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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 13:00, 17 November 2016

Problem

In the non-decreasing sequence of odd integers $\{a_1,a_2,a_3,\ldots \}=\{1,3,3,3,5,5,5,5,5,\ldots \}$ each odd positive integer $k$ appears $k$ times. It is a fact that there are integers $b, c$, and $d$ such that for all positive integers $n$, $a_n=b\lfloor \sqrt{n+c} \rfloor +d$, where $\lfloor x \rfloor$ denotes the largest integer not exceeding $x$. The sum $b+c+d$ equals

$\text{(A)} \ 0 \qquad  \text{(B)} \ 1 \qquad  \text{(C)} \ 2 \qquad  \text{(D)} \ 3 \qquad  \text{(E)} \ 4$

Solution

Solution by e_power_pi_times_i

Because the set consists of odd numbers, and since $\lfloor{}\sqrt{n+c}\rfloor{}$ is an integer and can be odd or even, $b = 2$ and $|a| = 1$. However, given that $\lfloor{}\sqrt{n+c}\rfloor{}$ can be $0$, $a = 1$. Then, $a_1 = 1 = 2\lfloor{}\sqrt{1+c}\rfloor{}+1$, and $\lfloor{}\sqrt{1+c}\rfloor{}$ = 0, and $c = -1$ because $c$ is an integer. $b+c+d = 2+(-1)+1 = \boxed{\text{(C)}\  2}$

See also

1980 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 24
Followed by
Problem 26
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